#include <bits/stdc++.h>
using namespace std;
const int mod = (int)(1e9) + 7;
int head[300005], v[300005 * 2], suc[300005 * 2], en = 1;
void addedge(int iu, int iv) {
  en++;
  v[en] = iv;
  suc[en] = head[iu];
  head[iu] = en;
}
long long V[300005], f[300005][2], g[300005][2];
long long ans;
void dfs1(int s, int par) {
  f[s][0] = g[s][0] = 0;
  f[s][1] = g[s][1] = 1;
  for (int i = head[s]; i; i = suc[i])
    if (v[i] != par) {
      dfs1(v[i], s);
      g[s][0] = (g[s][0] + (long long)f[s][0] * f[v[i]][1] % mod) % mod;
      g[s][1] = (g[s][1] + (long long)f[s][1] * f[v[i]][0] % mod) % mod;
      f[s][0] = (f[s][0] + f[v[i]][1]) % mod;
      f[s][1] = (f[s][1] + f[v[i]][0]) % mod;
    }
}
void dfs2(int s, int par, long long gp0, long long gp1) {
  ans = (ans +
         (long long)V[s] * (g[s][1] + (long long)f[s][1] * gp1 % mod) % mod) %
        mod;
  ans = (ans -
         (long long)V[s] * (g[s][0] + (long long)f[s][0] * gp0 % mod) % mod) %
        mod;
  ans = (ans % mod + mod) % mod;
  for (int i = head[s]; i; i = suc[i])
    if (v[i] != par) {
      dfs2(v[i], s, ((long long)gp1 + f[s][1] - f[v[i]][0]) % mod,
           ((long long)gp0 + f[s][0] - f[v[i]][1]) % mod);
    }
}
int main() {
  int i, j, k;
  int n;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%I64d", V + i);
  }
  for (i = 1; i < n; i++) {
    scanf("%d%d", &j, &k);
    addedge(j, k);
    addedge(k, j);
  }
  dfs1(1, 0);
  dfs2(1, 0, 0, 0);
  ans = 2ll * ans % mod;
  for (i = 1; i <= n; i++) ans = (ans - V[i]) % mod;
  ans = (ans % mod + mod) % mod;
  printf("%I64d", ans);
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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 mod = (int)(1e9) + 7;
int head[300005], v[300005 * 2], suc[300005 * 2], en = 1;
void addedge(int iu, int iv) {
  en++;
  v[en] = iv;
  suc[en] = head[iu];
  head[iu] = en;
}
long long V[300005], f[300005][2], g[300005][2];
long long ans;
void dfs1(int s, int par) {
  f[s][0] = g[s][0] = 0;
  f[s][1] = g[s][1] = 1;
  for (int i = head[s]; i; i = suc[i])
    if (v[i] != par) {
      dfs1(v[i], s);
      g[s][0] = (g[s][0] + (long long)f[s][0] * f[v[i]][1] % mod) % mod;
      g[s][1] = (g[s][1] + (long long)f[s][1] * f[v[i]][0] % mod) % mod;
      f[s][0] = (f[s][0] + f[v[i]][1]) % mod;
      f[s][1] = (f[s][1] + f[v[i]][0]) % mod;
    }
}
void dfs2(int s, int par, long long gp0, long long gp1) {
  ans = (ans +
         (long long)V[s] * (g[s][1] + (long long)f[s][1] * gp1 % mod) % mod) %
        mod;
  ans = (ans -
         (long long)V[s] * (g[s][0] + (long long)f[s][0] * gp0 % mod) % mod) %
        mod;
  ans = (ans % mod + mod) % mod;
  for (int i = head[s]; i; i = suc[i])
    if (v[i] != par) {
      dfs2(v[i], s, ((long long)gp1 + f[s][1] - f[v[i]][0]) % mod,
           ((long long)gp0 + f[s][0] - f[v[i]][1]) % mod);
    }
}
int main() {
  int i, j, k;
  int n;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%I64d", V + i);
  }
  for (i = 1; i < n; i++) {
    scanf("%d%d", &j, &k);
    addedge(j, k);
    addedge(k, j);
  }
  dfs1(1, 0);
  dfs2(1, 0, 0, 0);
  ans = 2ll * ans % mod;
  for (i = 1; i <= n; i++) ans = (ans - V[i]) % mod;
  ans = (ans % mod + mod) % mod;
  printf("%I64d", ans);
  return 0;
}

```