#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > adj(200001);
long long a[200001], c[200001][2], dp[200001][2], s[200001];
long long Solve(int i, int k) {
  long long ans = 0;
  for (int x = 0; x < 2; ++x) {
    for (int y = 0; y < 2; ++y) {
      long long a = x == 0 ? (1000000007 - dp[i][x]) % 1000000007 : dp[i][x];
      long long b = x == 0 ? dp[k][y] : (1000000007 - dp[k][y]) % 1000000007;
      long long ansx = (c[i][x] * b + c[k][y] * a) % 1000000007;
      ans = (ans + ansx) % 1000000007;
    }
  }
  return ans;
}
void DFS(int i, int p) {
  c[i][1] = 1;
  dp[i][1] = a[i];
  long long ans = 0;
  for (auto k : adj[i]) {
    if (k == p) continue;
    DFS(k, i);
    ans = (ans + Solve(i, k) + Solve(k, i)) % 1000000007;
    for (int x = 0; x < 2; ++x) {
      c[i][x] += c[k][1 - x];
      dp[i][x] = (dp[i][x] + c[k][1 - x] * a[i] + 1000000007 - dp[k][1 - x]) %
                 1000000007;
    }
  }
  s[i] = ans;
}
int main() {
  int n;
  scanf("%d", &n);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
    a[i] = (1000000007 + a[i]) % 1000000007;
    ans = (ans + a[i]) % 1000000007;
  }
  for (int x = 1; x < n; ++x) {
    int i, j;
    scanf("%d%d", &i, &j);
    adj[i].push_back(j);
    adj[j].push_back(i);
  }
  DFS(1, 0);
  for (int i = 1; i <= n; ++i) ans = (ans + s[i]) % 1000000007;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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;
vector<vector<int> > adj(200001);
long long a[200001], c[200001][2], dp[200001][2], s[200001];
long long Solve(int i, int k) {
  long long ans = 0;
  for (int x = 0; x < 2; ++x) {
    for (int y = 0; y < 2; ++y) {
      long long a = x == 0 ? (1000000007 - dp[i][x]) % 1000000007 : dp[i][x];
      long long b = x == 0 ? dp[k][y] : (1000000007 - dp[k][y]) % 1000000007;
      long long ansx = (c[i][x] * b + c[k][y] * a) % 1000000007;
      ans = (ans + ansx) % 1000000007;
    }
  }
  return ans;
}
void DFS(int i, int p) {
  c[i][1] = 1;
  dp[i][1] = a[i];
  long long ans = 0;
  for (auto k : adj[i]) {
    if (k == p) continue;
    DFS(k, i);
    ans = (ans + Solve(i, k) + Solve(k, i)) % 1000000007;
    for (int x = 0; x < 2; ++x) {
      c[i][x] += c[k][1 - x];
      dp[i][x] = (dp[i][x] + c[k][1 - x] * a[i] + 1000000007 - dp[k][1 - x]) %
                 1000000007;
    }
  }
  s[i] = ans;
}
int main() {
  int n;
  scanf("%d", &n);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
    a[i] = (1000000007 + a[i]) % 1000000007;
    ans = (ans + a[i]) % 1000000007;
  }
  for (int x = 1; x < n; ++x) {
    int i, j;
    scanf("%d%d", &i, &j);
    adj[i].push_back(j);
    adj[j].push_back(i);
  }
  DFS(1, 0);
  for (int i = 1; i <= n; ++i) ans = (ans + s[i]) % 1000000007;
  printf("%lld\n", ans);
  return 0;
}

```