E. Alternating Treetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputGiven a tree with n nodes numbered from 1 to n. Each node i has an associated value V_i.If the simple path from u_1 to u_m consists of m nodes namely u_1 \rightarrow u_2 \rightarrow u_3 \rightarrow \dots u_{m-1} \rightarrow u_{m}, then its alternating function A(u_{1},u_{m}) is defined as A(u_{1},u_{m}) = \sum\limits_{i=1}^{m} (-1)^{i+1} \cdot V_{u_{i}}. A path can also have 0 edges, i.e. u_{1}=u_{m}.Compute the sum of alternating functions of all unique simple paths. Note that the paths are directed: two paths are considered different if the starting vertices differ or the ending vertices differ. The answer may be large so compute it modulo 10^{9}+7. InputThe first line contains an integer n (2 \leq n \leq 2\cdot10^{5} ) — the number of vertices in the tree.The second line contains n space-separated integers V_1, V_2, \ldots, V_n (-10^9\leq V_i \leq 10^9) — values of the nodes.The next n-1 lines each contain two space-separated integers u and v (1\leq u, v\leq n, u \neq v) denoting an edge between vertices u and v. It is guaranteed that the given graph is a tree.OutputPrint the total sum of alternating functions of all unique simple paths modulo 10^{9}+7. ExamplesInput4-4 1 5 -21 21 31 4Output40Input8-2 6 -4 -4 -9 -3 -7 238 22 31 46 57 64 75 8Output4NoteConsider the first example.A simple path from node 1 to node 2: 1 \rightarrow 2 has alternating function equal to A(1,2) = 1 \cdot (-4)+(-1) \cdot 1 = -5.A simple path from node 1 to node 3: 1 \rightarrow 3 has alternating function equal to A(1,3) = 1 \cdot (-4)+(-1) \cdot 5 = -9.A simple path from node 2 to node 4: 2 \rightarrow 1 \rightarrow 4 has alternating function A(2,4) = 1 \cdot (1)+(-1) \cdot (-4)+1 \cdot (-2) = 3.A simple path from node 1 to node 1 has a single node 1, so A(1,1) = 1 \cdot (-4) = -4.Similarly, A(2, 1) = 5, A(3, 1) = 9, A(4, 2) = 3, A(1, 4) = -2, A(4, 1) = 2, A(2, 2) = 1, A(3, 3) = 5, A(4, 4) = -2, A(3, 4) = 7, A(4, 3) = 7, A(2, 3) = 10, A(3, 2) = 10. So the answer is (-5) + (-9) + 3 + (-4) + 5 + 9 + 3 + (-2) + 2 + 1 + 5 + (-2) + 7 + 7 + 10 + 10 = 40.Similarly A(1,4)=-2, A(2,2)=1, A(2,1)=5, A(2,3)=10, A(3,3)=5, A(3,1)=9, A(3,2)=10, A(3,4)=7, A(4,4)=-2, A(4,1)=2, A(4,2)=3 , A(4,3)=7 which sums upto 40. 
Input4-4 1 5 -21 21 31 4
Output40
['combinatorics', 'dfs and similar', 'divide and conquer', 'dp', 'probabilities', 'trees', '*2300']