#include <bits/stdc++.h>
using namespace std;
vector<long long> v[200005];
long long dp[200005];
long long sz[200005];
long long val[200005];
long long ans = 0;
long long dfs1(int cn, int pn) {
  sz[cn] = 1;
  for (auto x : v[cn]) {
    if (x != pn) {
      sz[cn] += dfs1(x, cn);
    }
  }
  return sz[cn];
}
long long dfs2(int cn, int pn) {
  dp[cn] = (sz[cn] * val[cn] + 1000000007) % 1000000007;
  for (auto x : v[cn]) {
    if (x != pn) {
      dp[cn] -= dfs2(x, cn);
      dp[cn] %= 1000000007;
      dp[cn] = (dp[cn] + 1000000007) % 1000000007;
    }
  }
  return dp[cn];
}
void dfs3(int cn, int pn) {
  ans = (ans + dp[cn] + 1000000007 + 1000000007) % 1000000007;
  for (auto x : v[cn]) {
    if (x != pn) {
      dp[cn] = (dp[cn] + dp[x] - sz[x] * val[cn] + 1000000007) % 1000000007;
      sz[cn] -= sz[x];
      dp[x] = (dp[x] + val[x] * sz[cn] - dp[cn] + 1000000007) % 1000000007;
      sz[x] += sz[cn];
      dfs3(x, cn);
      dp[x] = (dp[x] + dp[cn] - sz[cn] * val[x] + 1000000007) % 1000000007;
      sz[x] -= sz[cn];
      dp[cn] = (dp[cn] + val[cn] * sz[x] - dp[x] + 1000000007) % 1000000007;
      sz[cn] += sz[x];
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    long long n;
    cin >> n;
    for (long long i = 1; i < n + 1; i++) cin >> val[i];
    for (long long i = 0; i < n - 1; i++) {
      long long a, b;
      cin >> a >> b;
      v[a].push_back(b);
      v[b].push_back(a);
    }
    dfs1(1, 0);
    dfs2(1, 0);
    dfs3(1, 0);
    cout << ans << endl;
  }
  return 0;
}

### 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;
vector<long long> v[200005];
long long dp[200005];
long long sz[200005];
long long val[200005];
long long ans = 0;
long long dfs1(int cn, int pn) {
  sz[cn] = 1;
  for (auto x : v[cn]) {
    if (x != pn) {
      sz[cn] += dfs1(x, cn);
    }
  }
  return sz[cn];
}
long long dfs2(int cn, int pn) {
  dp[cn] = (sz[cn] * val[cn] + 1000000007) % 1000000007;
  for (auto x : v[cn]) {
    if (x != pn) {
      dp[cn] -= dfs2(x, cn);
      dp[cn] %= 1000000007;
      dp[cn] = (dp[cn] + 1000000007) % 1000000007;
    }
  }
  return dp[cn];
}
void dfs3(int cn, int pn) {
  ans = (ans + dp[cn] + 1000000007 + 1000000007) % 1000000007;
  for (auto x : v[cn]) {
    if (x != pn) {
      dp[cn] = (dp[cn] + dp[x] - sz[x] * val[cn] + 1000000007) % 1000000007;
      sz[cn] -= sz[x];
      dp[x] = (dp[x] + val[x] * sz[cn] - dp[cn] + 1000000007) % 1000000007;
      sz[x] += sz[cn];
      dfs3(x, cn);
      dp[x] = (dp[x] + dp[cn] - sz[cn] * val[x] + 1000000007) % 1000000007;
      sz[x] -= sz[cn];
      dp[cn] = (dp[cn] + val[cn] * sz[x] - dp[x] + 1000000007) % 1000000007;
      sz[cn] += sz[x];
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    long long n;
    cin >> n;
    for (long long i = 1; i < n + 1; i++) cin >> val[i];
    for (long long i = 0; i < n - 1; i++) {
      long long a, b;
      cin >> a >> b;
      v[a].push_back(b);
      v[b].push_back(a);
    }
    dfs1(1, 0);
    dfs2(1, 0);
    dfs3(1, 0);
    cout << ans << endl;
  }
  return 0;
}

```