#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int maxn = 2e5 + 10;
long long val[maxn], ret = 0;
vector<int> g[maxn];
int N, sz[maxn][2], colour[maxn];
void DFS1(int v, int p, int c) {
  colour[v] = c;
  sz[v][c] = 1;
  for (int u : g[v]) {
    if (u != p) {
      DFS1(u, v, 1 - c);
      for (int x : {0, 1}) {
        sz[v][x] += sz[u][x];
      }
    }
  }
  if (0)
    for (int x : {0, 1}) {
      printf("sz[%d][%d] = %d\n", v, x, sz[v][x]);
    }
}
void DFS2(int v, int p) {
  if (0) cout << "v = " << v << endl;
  ret += val[v];
  if (ret >= MOD) ret -= MOD;
  long long B = sz[1][colour[v]] - sz[v][colour[v]],
            W = sz[1][1 - colour[v]] - sz[v][1 - colour[v]];
  for (int u : g[v]) {
    if (u != p) {
      DFS2(u, v);
      ret += (val[v] * B % MOD) * (sz[u][0] + sz[u][1]) % MOD;
      if (ret >= MOD) ret -= MOD;
      ret += (val[v] * sz[u][colour[v]] % MOD) * (B + W) % MOD;
      if (ret >= MOD) ret -= MOD;
      ret -= (val[v] * sz[u][1 - colour[v]] % MOD) * (B + W) % MOD;
      if (ret < 0) ret += MOD;
      ret -= (val[v] * W % MOD) * (sz[u][0] + sz[u][1]) % MOD;
      if (ret < 0) ret += MOD;
      B += sz[u][colour[v]];
      W += sz[u][1 - colour[v]];
    }
  }
  ret += (val[v] * B % MOD) * 2 % MOD;
  if (ret >= MOD) ret -= MOD;
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) {
    scanf("%lld", &val[i]);
  }
  for (int i = 1; i < N; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  DFS1(1, 1, 1);
  DFS2(1, 1);
  printf("%lld\n", (ret % MOD + MOD) % MOD);
}

### Prompt

Please provide a cpp coded solution 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 MOD = 1e9 + 7;
const int maxn = 2e5 + 10;
long long val[maxn], ret = 0;
vector<int> g[maxn];
int N, sz[maxn][2], colour[maxn];
void DFS1(int v, int p, int c) {
  colour[v] = c;
  sz[v][c] = 1;
  for (int u : g[v]) {
    if (u != p) {
      DFS1(u, v, 1 - c);
      for (int x : {0, 1}) {
        sz[v][x] += sz[u][x];
      }
    }
  }
  if (0)
    for (int x : {0, 1}) {
      printf("sz[%d][%d] = %d\n", v, x, sz[v][x]);
    }
}
void DFS2(int v, int p) {
  if (0) cout << "v = " << v << endl;
  ret += val[v];
  if (ret >= MOD) ret -= MOD;
  long long B = sz[1][colour[v]] - sz[v][colour[v]],
            W = sz[1][1 - colour[v]] - sz[v][1 - colour[v]];
  for (int u : g[v]) {
    if (u != p) {
      DFS2(u, v);
      ret += (val[v] * B % MOD) * (sz[u][0] + sz[u][1]) % MOD;
      if (ret >= MOD) ret -= MOD;
      ret += (val[v] * sz[u][colour[v]] % MOD) * (B + W) % MOD;
      if (ret >= MOD) ret -= MOD;
      ret -= (val[v] * sz[u][1 - colour[v]] % MOD) * (B + W) % MOD;
      if (ret < 0) ret += MOD;
      ret -= (val[v] * W % MOD) * (sz[u][0] + sz[u][1]) % MOD;
      if (ret < 0) ret += MOD;
      B += sz[u][colour[v]];
      W += sz[u][1 - colour[v]];
    }
  }
  ret += (val[v] * B % MOD) * 2 % MOD;
  if (ret >= MOD) ret -= MOD;
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) {
    scanf("%lld", &val[i]);
  }
  for (int i = 1; i < N; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  DFS1(1, 1, 1);
  DFS2(1, 1);
  printf("%lld\n", (ret % MOD + MOD) % MOD);
}

```