#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N = 200000 + 10, MOD = 1e9 + 7;
int n, v[N];
int ans = 0;
vector<int> G[N];
int size[N][2];
void dfs_size(int x, int fa) {
  size[x][0] = 1;
  size[x][1] = 0;
  for (int v : G[x])
    if (v != fa) {
      dfs_size(v, x);
      size[x][0] += size[v][1];
      size[x][1] += size[v][0];
    }
}
inline int norm(int x) {
  if (x >= MOD || x <= -MOD) x %= MOD;
  if (x < 0) x += MOD;
  return x;
}
inline int add(int x, int y) {
  x = norm(x);
  y = norm(y);
  return (x + y) % MOD;
}
void dfs(int x, int fa, int up0, int up1, int up_size) {
  int sizer = size[x][0] + size[x][1];
  ans = add(ans, 1ll * (up0 + 1) * sizer % MOD * v[x] % MOD);
  ans = add(ans, -1ll * up1 * sizer % MOD * v[x] % MOD);
  ans = add(ans, 1ll * size[x][0] * (up_size + 1) % MOD * v[x] % MOD);
  ans = add(ans, -1ll * size[x][1] * (up_size + 1) % MOD * v[x] % MOD);
  ans = add(ans, -v[x]);
  for (int t : G[x])
    if (t != fa) {
      int sizet = size[t][0] + size[t][1];
      ans =
          add(ans, -1ll * size[t][0] * (sizer - sizet - 1) % MOD * v[x] % MOD);
      ans = add(ans, 1ll * size[t][1] * (sizer - sizet - 1) % MOD * v[x] % MOD);
    }
  for (int v : G[x])
    if (v != fa)
      dfs(v, x, up1 + size[x][1] - size[v][0], up0 + size[x][0] - size[v][1],
          up_size + sizer - size[v][0] - size[v][1]);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs_size(1, -1);
  dfs(1, -1, 0, 0, 0);
  printf("%d\n", ans);
}

### Prompt

Create a solution in CPP for 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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N = 200000 + 10, MOD = 1e9 + 7;
int n, v[N];
int ans = 0;
vector<int> G[N];
int size[N][2];
void dfs_size(int x, int fa) {
  size[x][0] = 1;
  size[x][1] = 0;
  for (int v : G[x])
    if (v != fa) {
      dfs_size(v, x);
      size[x][0] += size[v][1];
      size[x][1] += size[v][0];
    }
}
inline int norm(int x) {
  if (x >= MOD || x <= -MOD) x %= MOD;
  if (x < 0) x += MOD;
  return x;
}
inline int add(int x, int y) {
  x = norm(x);
  y = norm(y);
  return (x + y) % MOD;
}
void dfs(int x, int fa, int up0, int up1, int up_size) {
  int sizer = size[x][0] + size[x][1];
  ans = add(ans, 1ll * (up0 + 1) * sizer % MOD * v[x] % MOD);
  ans = add(ans, -1ll * up1 * sizer % MOD * v[x] % MOD);
  ans = add(ans, 1ll * size[x][0] * (up_size + 1) % MOD * v[x] % MOD);
  ans = add(ans, -1ll * size[x][1] * (up_size + 1) % MOD * v[x] % MOD);
  ans = add(ans, -v[x]);
  for (int t : G[x])
    if (t != fa) {
      int sizet = size[t][0] + size[t][1];
      ans =
          add(ans, -1ll * size[t][0] * (sizer - sizet - 1) % MOD * v[x] % MOD);
      ans = add(ans, 1ll * size[t][1] * (sizer - sizet - 1) % MOD * v[x] % MOD);
    }
  for (int v : G[x])
    if (v != fa)
      dfs(v, x, up1 + size[x][1] - size[v][0], up0 + size[x][0] - size[v][1],
          up_size + sizer - size[v][0] - size[v][1]);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs_size(1, -1);
  dfs(1, -1, 0, 0, 0);
  printf("%d\n", ans);
}

```