CF960E.Alternating Tree

Given 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 $10^{9}+7$ .

The 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.

Print the total sum of alternating functions of all unique simple paths modulo $10^{9}+7$ .

Consider 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.