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 $10^{9}+7$.

给定一棵包含 $n$ 个节点的树，节点编号为 $1$ 到 $n$。每个节点 $i$ 关联一个值 $V_i$。

若从 $u_1$ 到 $u_m$ 的简单路径由 $m$ 个节点组成，即 $u_1 \rightarrow u_2 \rightarrow u_3 \rightarrow \dots \rightarrow u_{m-1} \rightarrow u_{m}$，则该路径的交替函数 $A(u_{1},u_{m})$ 定义为

$$A(u_{1},u_{m}) = \sum\limits_{i=1}^{m} (-1)^{i+1} \cdot V_{u_{i}}.$$

路径也可以包含 $0$ 条边，即 $u_1 = u_m$。

请计算所有不同的简单路径的交替函数之和。注意：路径是有向的——若两条路径的起点不同或终点不同，则视为不同路径。答案可能很大，请对 $10^{9}+7$ 取模。

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.

第一行包含一个整数 $n$（$2 \leq n \leq 2\cdot10^{5}$）——树中顶点的数量。

第二行包含 $n$ 个用空格分隔的整数 $V_1, V_2, \ldots, V_n$（$-10^9\leq V_i \leq 10^9$）——各节点的值。

接下来的 $n-1$ 行，每行包含两个用空格分隔的整数 $u$ 和 $v$（$1\leq u, v\leq n, u \neq v$），表示顶点 $u$ 与 $v$ 之间存在一条边。保证所给图是一棵树。

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

输出所有唯一简单路径的交替函数的总和对 $10^{9}+7$ 取模的结果。

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.

考虑第一个例子。

从节点 $1$ 到节点 $2$ 的一条简单路径：$1 \rightarrow 2$，其交替函数值为 $A(1,2) = 1 \cdot (-4)+(-1) \cdot 1 = -5$。

从节点 $1$ 到节点 $3$ 的一条简单路径：$1 \rightarrow 3$，其交替函数值为 $A(1,3) = 1 \cdot (-4)+(-1) \cdot 5 = -9$。

从节点 $2$ 到节点 $4$ 的一条简单路径：$2 \rightarrow 1 \rightarrow 4$，其交替函数值为 $A(2,4) = 1 \cdot (1)+(-1) \cdot (-4)+1 \cdot (-2) = 3$。

从节点 $1$ 到节点 $1$ 的一条简单路径仅含单个节点 $1$，因此 $A(1,1) = 1 \cdot (-4) = -4$。

类似地，$A(2, 1) = 5$，$A(3, 1) = 9$，$A(4, 2) = 3$，$A(1, 4) = -2$，$A(4, 1) = 2$，$A(2, 2) = 1$，$A(3, 3) = 5$，$A(4, 4) = -2$，$A(3, 4) = 7$，$A(4, 3) = 7$，$A(2, 3) = 10$，$A(3, 2) = 10$。因此答案为 $(-5) + (-9) + 3 + (-4) + 5 + 9 + 3 + (-2) + 2 + 1 + 5 + (-2) + 7 + 7 + 10 + 10 = 40$。

类似地，$A(1,4)=-2,\ A(2,2)=1,\ A(2,1)=5,\ A(2,3)=10,\ A(3,3)=5,\ A(3,1)=9,\ A(3,2)=10,\ A(3,4)=7,\ A(4,4)=-2,\ A(4,1)=2,\ A(4,2)=3,\ A(4,3)=7$，其和也为 40。