| <p> | |
| A group of <strong>N</strong> Foxen reside in a peaceful forest community. | |
| Each Fox's property consists of a tree stump as well as an underground den. | |
| There are <strong>N</strong> - 1 two-way paths on the ground running amongst the tree stumps, | |
| with the <em>i</em>th path connecting the stumps belonging to two different Foxen | |
| <strong>A<sub>i</sub></strong> and <strong>B<sub>i</sub></strong>, | |
| such that all <strong>N</strong> stumps can be reached from one another by following a sequence of paths. | |
| Similarly, there are <strong>N</strong> - 1 underground tunnels running amongst the dens, | |
| with the <em>i</em>th tunnel connecting the dens belonging to Foxen | |
| <strong>C<sub>i</sub></strong> and <strong>D<sub>i</sub></strong>, | |
| such that all <strong>N</strong> dens can be reached from one another. | |
| There's additionally a passageway connecting the tree stump and den belonging to the 1st Fox, | |
| which is the only way in the whole forest to get underground from the surface and vice versa. | |
| </p> | |
| <p> | |
| At night the Foxen sleep in their dens, but during the daytime, they like to emerge and relax lazily on their tree stumps. | |
| Each day, every Fox takes a trip from their den to their tree stump, | |
| taking the unique shortest path through the system of tunnels and paths to get there. | |
| However, this often requires passing through other Foxen's properties, which they don't appreciate a whole lot. | |
| To compensate, the Foxen have started charging each other tolls for said passage. | |
| They don't have much of a currency, but Foxen do love crackers, so those will do. | |
| Over a given period of <strong>M</strong> days, on the <em>i</em>th day, two different Foxen | |
| <strong>W<sub>i</sub></strong> and <strong>X<sub>i</sub></strong> will each charge tolls for one of their pieces of property. | |
| If <strong>Y<sub>i</sub></strong> = "T", | |
| then Fox <strong>W<sub>i</sub></strong> will be charging tolls for passage through their tree stump. | |
| Otherwise, if <strong>Y<sub>i</sub></strong> = "D", | |
| then Fox <strong>W<sub>i</sub></strong> will instead be charging tolls for passage through their den. | |
| Similarly, Fox <strong>X<sub>i</sub></strong> will be charging tolls for passage through either their tree stump | |
| (if <strong>Z<sub>i</sub></strong>= "T") or their den (if <strong>Z<sub>i</sub></strong>= "D"). | |
| </p> | |
| <p> | |
| Each day, whenever a Fox passes through another Fox's den or stump which is subject to tolls on that day, | |
| they'll normally need to pay up with 2 crackers. However, if they've already paid a toll earlier on that same trip, | |
| then the property-owning Fox will take pity and only charge them 1 cracker instead of 2. | |
| As such, a Fox's daily trip may end up costing them at most 3 crackers. A Fox will never charge themselves a toll, of course. If a pair of Foxen both owe each other crackers, they'll still both pay up as normal, rather than attempting to minimize the number of cracker transactions performed. | |
| </p> | |
| <p> | |
| The Foxen are having some trouble keeping track of how many crackers they owe one another. On each of the <strong>M</strong> days, they'd like to count up the total number of crackers which will be charged as part of the tolls for the <strong>N</strong> trips taken on that day. To avoid dealing with too many large numbers, they'd like to combine these <strong>M</strong> cracker counts into a single value as follows (where <strong>V<sub>i</sub></strong> is the <em>i</em>th day's count): | |
| </p> | |
| <p> | |
| | |
| ( ... (((<strong>V<sub>1</sub></strong> * 12,345) + <strong>V<sub>2</sub></strong>) * 12,345 + <strong>V<sub>3</sub></strong>) ... * 12,345 + <strong>V<sub>M</sub></strong>) modulo 1,000,000,007 | |
| </p> | |
| <p> | |
| Please help the Foxen compute this combined value! | |
| </p> | |
| <h3>Input</h3> | |
| <p> | |
| Input begins with an integer <strong>T</strong>, the number of different communities of Foxen. | |
| For each community of Foxen, there is first a line containing the space-separated integers <strong>N</strong> and <strong>M</strong>. | |
| Then <strong>N - 1</strong> lines follow, the <em>i</em>th of which contains the space-separated integers | |
| <strong>A<sub>i</sub></strong> and <strong>B<sub>i</sub></strong>. | |
| Then <strong>N - 1</strong> lines follow, the <em>i</em>th of which contains the space-separated integers | |
| <strong>C<sub>i</sub></strong> and <strong>D<sub>i</sub></strong>. | |
| Then <strong>M</strong> lines follow, the <em>i</em>th of which contains the integers | |
| <strong>W<sub>i</sub></strong> and <strong>X<sub>i</sub></strong> and the characters | |
| <strong>Y<sub>i</sub></strong> and <strong>Z<sub>i</sub></strong>, all separated by spaces. | |
| </p> | |
| <h3>Output</h3> | |
| <p> | |
| For the <em>i</em>th community of Foxen, print a line containing "Case #<strong>i</strong>: " | |
| followed by a single integer, the requested combined value based on the <strong>M</strong> days' cracker counts, modulo 1,000,000,007. | |
| </p> | |
| <h3>Constraints</h3> | |
| <p> | |
| 1 ≤ <strong>T</strong> ≤ 30<br /> | |
| 2 ≤ <strong>N</strong> ≤ 500,000 <br /> | |
| 1 ≤ <strong>M</strong> ≤ 500,000 <br /> | |
| 1 ≤ | |
| <strong>A<sub>i</sub></strong>, <strong>B<sub>i</sub></strong>, | |
| <strong>C<sub>i</sub></strong>, <strong>D<sub>i</sub></strong>, | |
| <strong>W<sub>i</sub></strong>, <strong>X<sub>i</sub></strong> | |
| ≤ <strong>N</strong> <br /> | |
| Both the sum of <strong>N</strong> values and the sum of <strong>M</strong> values across all <strong>T</strong> cases do not exceed 1,500,000. | |
| </p> | |
| <h3>Explanation of Sample</h3> | |
| <p> | |
| In the first case, Fox 1 doesn't need to pay any tolls to get from its den to its tree stump, while Fox 2 must pay 2 crackers to complete its trip due to passing through Fox's 1 tree stump. | |
| </p> | |
| <p> | |
| In the second case, 5 crackers will be charged on the first day (the 3 Foxen must pay 0, 2, and 3 crackers, respectively), 2 crackers will be charged on the second day, and none will be charged on the third day. This results in a final answer of (((5 * 12,345) + 2) * 12,345) + 0) modulo 1,000,000,007 = 762,019,815. | |
| </p> | |
| <p> | |
| In the third case, 7, 6, and 4 crackers will be charged on each of the three days, respectively. | |
| </p> | |