#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int NMAX = 200005;
int v[NMAX], h[NMAX], fr[3], howmany[NMAX][2];
vector<int> g[NMAX];
void dfs(int from, int x) {
  h[from] = x;
  fr[x % 2]++;
  howmany[from][x % 2] = 1;
  for (auto it : g[from])
    if (h[it] == 0) {
      dfs(it, x + 1);
      howmany[from][0] += howmany[it][0];
      howmany[from][1] += howmany[it][1];
    }
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    if (v[i] < 0) v[i] = MOD + v[i];
  }
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int up = fr[h[i] % 2] - howmany[i][h[i] % 2];
    int down = howmany[i][0] + howmany[i][1];
    ans = (ans + 1LL * ((1LL * v[i] * up) % MOD) * down) % MOD;
    down = howmany[i][h[i] % 2] - 1;
    up = howmany[1][0] + howmany[1][1] - howmany[i][1] - howmany[i][0] + 1;
    ans = (ans + 1LL * ((1LL * v[i] * up) % MOD) * down) % MOD;
    for (auto it : g[i]) {
      if (h[it] == h[i] + 1) {
        int from = howmany[it][h[i] % 2];
        int to =
            howmany[i][0] + howmany[i][1] - howmany[it][0] - howmany[it][1] - 1;
        ans = (ans + 1LL * ((1LL * v[i] * from) % MOD) * to) % MOD;
      }
    }
    ans = (ans + 1LL * v[i] * n) % MOD;
  }
  for (int i = 1; i <= n; i++) {
    int up = fr[(h[i] + 1) % 2] - howmany[i][(h[i] + 1) % 2];
    int down = howmany[i][0] + howmany[i][1];
    ans = (ans + MOD - (1LL * ((1LL * v[i] * up) % MOD) * down) % MOD) % MOD;
    down = howmany[i][(h[i] + 1) % 2];
    up = howmany[1][0] + howmany[1][1] - howmany[i][1] - howmany[i][0] + 1;
    ans = (ans + MOD - (1LL * ((1LL * v[i] * up) % MOD) * down) % MOD) % MOD;
    for (auto it : g[i]) {
      if (h[it] == h[i] + 1) {
        int from = howmany[it][(h[i] + 1) % 2];
        int to =
            howmany[i][0] + howmany[i][1] - howmany[it][0] - howmany[it][1] - 1;
        ans =
            (ans + MOD - (1LL * ((1LL * v[i] * from) % MOD) * to) % MOD) % MOD;
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 = 1000000007;
const int NMAX = 200005;
int v[NMAX], h[NMAX], fr[3], howmany[NMAX][2];
vector<int> g[NMAX];
void dfs(int from, int x) {
  h[from] = x;
  fr[x % 2]++;
  howmany[from][x % 2] = 1;
  for (auto it : g[from])
    if (h[it] == 0) {
      dfs(it, x + 1);
      howmany[from][0] += howmany[it][0];
      howmany[from][1] += howmany[it][1];
    }
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    if (v[i] < 0) v[i] = MOD + v[i];
  }
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int up = fr[h[i] % 2] - howmany[i][h[i] % 2];
    int down = howmany[i][0] + howmany[i][1];
    ans = (ans + 1LL * ((1LL * v[i] * up) % MOD) * down) % MOD;
    down = howmany[i][h[i] % 2] - 1;
    up = howmany[1][0] + howmany[1][1] - howmany[i][1] - howmany[i][0] + 1;
    ans = (ans + 1LL * ((1LL * v[i] * up) % MOD) * down) % MOD;
    for (auto it : g[i]) {
      if (h[it] == h[i] + 1) {
        int from = howmany[it][h[i] % 2];
        int to =
            howmany[i][0] + howmany[i][1] - howmany[it][0] - howmany[it][1] - 1;
        ans = (ans + 1LL * ((1LL * v[i] * from) % MOD) * to) % MOD;
      }
    }
    ans = (ans + 1LL * v[i] * n) % MOD;
  }
  for (int i = 1; i <= n; i++) {
    int up = fr[(h[i] + 1) % 2] - howmany[i][(h[i] + 1) % 2];
    int down = howmany[i][0] + howmany[i][1];
    ans = (ans + MOD - (1LL * ((1LL * v[i] * up) % MOD) * down) % MOD) % MOD;
    down = howmany[i][(h[i] + 1) % 2];
    up = howmany[1][0] + howmany[1][1] - howmany[i][1] - howmany[i][0] + 1;
    ans = (ans + MOD - (1LL * ((1LL * v[i] * up) % MOD) * down) % MOD) % MOD;
    for (auto it : g[i]) {
      if (h[it] == h[i] + 1) {
        int from = howmany[it][(h[i] + 1) % 2];
        int to =
            howmany[i][0] + howmany[i][1] - howmany[it][0] - howmany[it][1] - 1;
        ans =
            (ans + MOD - (1LL * ((1LL * v[i] * from) % MOD) * to) % MOD) % MOD;
      }
    }
  }
  cout << ans;
  return 0;
}

```