#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, mo = 1e9 + 7;
int n, x, y, js, la[N], ne[N * 2], to[N * 2], v[N], p[N];
long long ans, an[N][4], g[N][4], aa;
void R(int &n) {
  char c;
  int fh = 1;
  for (; (c = getchar()) < '0' || c > '9';)
    if (c == '-') fh = -1;
  for (n = 0; c <= '9' && c >= '0'; c = getchar()) n = n * 10 + c - 48;
  n *= fh;
}
void dfs(int x) {
  int y;
  for (int i = la[x]; i; i = ne[i])
    if ((y = to[i]) != p[x]) {
      p[y] = x;
      dfs(y);
      an[x][1] = (g[y][0] * (mo - v[x]) + an[y][0] + an[x][1]) % mo;
      g[x][1] += g[y][0];
      an[x][0] = (g[y][1] * v[x] + an[y][1] + an[x][0]) % mo;
      g[x][0] += g[y][1];
      an[x][3] = (an[x][3] + g[y][2] * (mo - v[x]) + an[y][2]) % mo;
      g[x][3] += g[y][2];
      an[x][2] = (an[x][2] + (long long)v[x] * g[y][3] + an[y][3]) % mo;
      g[x][2] += g[y][3];
    }
  ans = (ans + an[x][0] + an[x][1] + an[x][2]) % mo;
  aa = 0;
  for (int i = la[x]; i; i = ne[i])
    if ((y = to[i]) != p[x]) {
      aa =
          (aa +
           (an[x][0] - g[y][1] * v[x] % mo - an[y][1] + mo * 2) % mo * g[y][3] +
           an[y][3] * (g[x][0] - g[y][1]) +
           (an[x][1] - g[y][0] * (mo - v[x]) % mo - an[y][0] + mo * 2) % mo *
               g[y][2] +
           an[y][2] * (g[x][1] - g[y][0])) %
          mo;
      aa =
          (aa +
           (an[x][2] - g[y][3] * v[x] % mo - an[y][3] + mo * 2) % mo * g[y][1] +
           an[y][1] * (g[x][2] - g[y][3]) +
           (an[x][3] - g[y][2] * (mo - v[x]) % mo - an[y][2] + mo * 2) % mo *
               g[y][0] +
           an[y][0] * (g[x][3] - g[y][2])) %
          mo;
    }
  ans = (ans + aa * (mo + 1 >> 1) + v[x]) % mo;
  an[x][0] = (an[x][0] + v[x]) % mo;
  g[x][0]++;
  an[x][2] = (an[x][2] + v[x]) % mo;
  g[x][2]++;
  an[x][3] = (an[x][3] + mo - v[x]) % mo;
  g[x][3]++;
}
int main() {
  R(n);
  for (int i = 1, E = n; i <= E; i++) R(v[i]), v[i] = (v[i] + mo) % mo;
  for (int i = 2, E = n; i <= E; i++)
    R(x), R(y), (ne[++js] = la[x], la[x] = js, to[js] = y),
        (ne[++js] = la[y], la[y] = js, to[js] = x);
  dfs(1);
  printf("%I64d", ans);
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 → u_2 → u_3 → ... u_{m-1} → u_{m}, then its alternating function A(u_{1},u_{m}) is defined as A(u_{1},u_{m}) = ∑_{i=1}^{m} (-1)^{i+1} ⋅ 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. 

Input

The first line contains an integer n (2 ≤ n ≤ 2⋅10^{5} ) — the number of vertices in the tree.

The second line contains n space-separated integers V_1, V_2, …, V_n (-10^9≤ V_i ≤ 10^9) — values of the nodes.

The next n-1 lines each contain two space-separated integers u and v (1≤ u, v≤ n, u ≠ v) denoting an edge between vertices u and v. It is guaranteed that the given graph is a tree.

Output

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

Examples

Input

4
-4 1 5 -2
1 2
1 3
1 4


Output

40


Input

8
-2 6 -4 -4 -9 -3 -7 23
8 2
2 3
1 4
6 5
7 6
4 7
5 8


Output

4

Note

Consider the first example.

A simple path from node 1 to node 2: 1 → 2 has alternating function equal to A(1,2) = 1 ⋅ (-4)+(-1) ⋅ 1 = -5.

A simple path from node 1 to node 3: 1 → 3 has alternating function equal to A(1,3) = 1 ⋅ (-4)+(-1) ⋅ 5 = -9.

A simple path from node 2 to node 4: 2 → 1 → 4 has alternating function A(2,4) = 1 ⋅ (1)+(-1) ⋅ (-4)+1 ⋅ (-2) = 3.

A simple path from node 1 to node 1 has a single node 1, so A(1,1) = 1 ⋅ (-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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, mo = 1e9 + 7;
int n, x, y, js, la[N], ne[N * 2], to[N * 2], v[N], p[N];
long long ans, an[N][4], g[N][4], aa;
void R(int &n) {
  char c;
  int fh = 1;
  for (; (c = getchar()) < '0' || c > '9';)
    if (c == '-') fh = -1;
  for (n = 0; c <= '9' && c >= '0'; c = getchar()) n = n * 10 + c - 48;
  n *= fh;
}
void dfs(int x) {
  int y;
  for (int i = la[x]; i; i = ne[i])
    if ((y = to[i]) != p[x]) {
      p[y] = x;
      dfs(y);
      an[x][1] = (g[y][0] * (mo - v[x]) + an[y][0] + an[x][1]) % mo;
      g[x][1] += g[y][0];
      an[x][0] = (g[y][1] * v[x] + an[y][1] + an[x][0]) % mo;
      g[x][0] += g[y][1];
      an[x][3] = (an[x][3] + g[y][2] * (mo - v[x]) + an[y][2]) % mo;
      g[x][3] += g[y][2];
      an[x][2] = (an[x][2] + (long long)v[x] * g[y][3] + an[y][3]) % mo;
      g[x][2] += g[y][3];
    }
  ans = (ans + an[x][0] + an[x][1] + an[x][2]) % mo;
  aa = 0;
  for (int i = la[x]; i; i = ne[i])
    if ((y = to[i]) != p[x]) {
      aa =
          (aa +
           (an[x][0] - g[y][1] * v[x] % mo - an[y][1] + mo * 2) % mo * g[y][3] +
           an[y][3] * (g[x][0] - g[y][1]) +
           (an[x][1] - g[y][0] * (mo - v[x]) % mo - an[y][0] + mo * 2) % mo *
               g[y][2] +
           an[y][2] * (g[x][1] - g[y][0])) %
          mo;
      aa =
          (aa +
           (an[x][2] - g[y][3] * v[x] % mo - an[y][3] + mo * 2) % mo * g[y][1] +
           an[y][1] * (g[x][2] - g[y][3]) +
           (an[x][3] - g[y][2] * (mo - v[x]) % mo - an[y][2] + mo * 2) % mo *
               g[y][0] +
           an[y][0] * (g[x][3] - g[y][2])) %
          mo;
    }
  ans = (ans + aa * (mo + 1 >> 1) + v[x]) % mo;
  an[x][0] = (an[x][0] + v[x]) % mo;
  g[x][0]++;
  an[x][2] = (an[x][2] + v[x]) % mo;
  g[x][2]++;
  an[x][3] = (an[x][3] + mo - v[x]) % mo;
  g[x][3]++;
}
int main() {
  R(n);
  for (int i = 1, E = n; i <= E; i++) R(v[i]), v[i] = (v[i] + mo) % mo;
  for (int i = 2, E = n; i <= E; i++)
    R(x), R(y), (ne[++js] = la[x], la[x] = js, to[js] = y),
        (ne[++js] = la[y], la[y] = js, to[js] = x);
  dfs(1);
  printf("%I64d", ans);
}

```