#include <bits/stdc++.h>
using namespace std;
long long cd, v[200001], even_up[200001], odd_up[200001], odd_left[200001],
    odd_right[200001], even_left[200001], even_right[200001];
long long even_down[200001], odd_down[200001], sz[200001];
long long result = 0, mod = 1e9 + 7;
vector<long long> edge[200001];
void inp() {
  long long a, b;
  cin >> cd;
  for (long long i = 1; i <= cd; i++) cin >> v[i];
  for (long long i = 1; i < cd; i++)
    cin >> a >> b, edge[a].push_back(b), edge[b].push_back(a);
}
void dfs_up(long long u, long long pre) {
  odd_up[u] = 1;
  for (auto v : edge[u]) {
    if (v != pre) {
      dfs_up(v, u);
      even_up[u] += odd_up[v];
      odd_up[u] += even_up[v];
    }
  }
}
void dfs_lr(long long u, long long pre) {
  int previous = -1;
  for (long long i = 0; i < edge[u].size(); i++) {
    if (edge[u][i] == pre) continue;
    odd_left[edge[u][i]] = odd_up[edge[u][i]];
    even_left[edge[u][i]] = even_up[edge[u][i]];
    if (previous != -1)
      odd_left[edge[u][i]] += odd_left[previous],
          even_left[edge[u][i]] += even_left[previous];
    previous = edge[u][i];
  }
  previous = -1;
  for (long long i = edge[u].size() - 1; i >= 0; i--) {
    if (edge[u][i] == pre) continue;
    odd_right[edge[u][i]] = odd_up[edge[u][i]];
    even_right[edge[u][i]] = even_up[edge[u][i]];
    if (i != edge[u].size() - 1)
      odd_right[edge[u][i]] += odd_right[previous],
          even_right[edge[u][i]] += even_right[previous];
    previous = edge[u][i];
  }
  for (auto v : edge[u])
    if (v != pre) dfs_lr(v, u);
}
void dfs_down(long long u, long long pre) {
  sz[u] = 1;
  for (auto v : edge[u]) {
    if (v != pre) {
      odd_down[v] =
          even_down[u] + odd_left[v] - 2 * odd_up[v] + odd_right[v] + 1;
      even_down[v] =
          odd_down[u] + even_left[v] - 2 * even_up[v] + even_right[v];
      dfs_down(v, u), sz[u] += sz[v];
    }
  }
}
void dfs_result(int u, int pre) {
  for (auto v1 : edge[u]) {
    if (v1 == pre) continue;
    int remain = sz[u] - sz[v1];
    result += (even_up[v1] * remain % mod * v[u]) % mod;
    result %= mod;
    result -= (odd_up[v1] * remain % mod * v[u]) % mod;
    result += mod * mod;
    result %= mod;
    dfs_result(v1, u);
  }
}
void do_it() {
  dfs_up(1, 0);
  dfs_lr(1, 0);
  odd_down[1] = 1;
  dfs_down(1, 0);
  dfs_result(1, 0);
  for (long long i = 1; i <= cd; i++) {
    result +=
        ((odd_down[i] * sz[i] % mod + odd_up[i] * (cd - sz[i])) % mod * v[i]) %
        mod;
    result %= mod;
    result -= ((even_down[i] * sz[i] % mod + even_up[i] * (cd - sz[i])) % mod *
               v[i]) %
              mod;
    result += mod * mod;
    result %= mod;
  }
  cout << result << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  inp();
  do_it();
}

### Prompt

Develop a solution in cpp 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;
long long cd, v[200001], even_up[200001], odd_up[200001], odd_left[200001],
    odd_right[200001], even_left[200001], even_right[200001];
long long even_down[200001], odd_down[200001], sz[200001];
long long result = 0, mod = 1e9 + 7;
vector<long long> edge[200001];
void inp() {
  long long a, b;
  cin >> cd;
  for (long long i = 1; i <= cd; i++) cin >> v[i];
  for (long long i = 1; i < cd; i++)
    cin >> a >> b, edge[a].push_back(b), edge[b].push_back(a);
}
void dfs_up(long long u, long long pre) {
  odd_up[u] = 1;
  for (auto v : edge[u]) {
    if (v != pre) {
      dfs_up(v, u);
      even_up[u] += odd_up[v];
      odd_up[u] += even_up[v];
    }
  }
}
void dfs_lr(long long u, long long pre) {
  int previous = -1;
  for (long long i = 0; i < edge[u].size(); i++) {
    if (edge[u][i] == pre) continue;
    odd_left[edge[u][i]] = odd_up[edge[u][i]];
    even_left[edge[u][i]] = even_up[edge[u][i]];
    if (previous != -1)
      odd_left[edge[u][i]] += odd_left[previous],
          even_left[edge[u][i]] += even_left[previous];
    previous = edge[u][i];
  }
  previous = -1;
  for (long long i = edge[u].size() - 1; i >= 0; i--) {
    if (edge[u][i] == pre) continue;
    odd_right[edge[u][i]] = odd_up[edge[u][i]];
    even_right[edge[u][i]] = even_up[edge[u][i]];
    if (i != edge[u].size() - 1)
      odd_right[edge[u][i]] += odd_right[previous],
          even_right[edge[u][i]] += even_right[previous];
    previous = edge[u][i];
  }
  for (auto v : edge[u])
    if (v != pre) dfs_lr(v, u);
}
void dfs_down(long long u, long long pre) {
  sz[u] = 1;
  for (auto v : edge[u]) {
    if (v != pre) {
      odd_down[v] =
          even_down[u] + odd_left[v] - 2 * odd_up[v] + odd_right[v] + 1;
      even_down[v] =
          odd_down[u] + even_left[v] - 2 * even_up[v] + even_right[v];
      dfs_down(v, u), sz[u] += sz[v];
    }
  }
}
void dfs_result(int u, int pre) {
  for (auto v1 : edge[u]) {
    if (v1 == pre) continue;
    int remain = sz[u] - sz[v1];
    result += (even_up[v1] * remain % mod * v[u]) % mod;
    result %= mod;
    result -= (odd_up[v1] * remain % mod * v[u]) % mod;
    result += mod * mod;
    result %= mod;
    dfs_result(v1, u);
  }
}
void do_it() {
  dfs_up(1, 0);
  dfs_lr(1, 0);
  odd_down[1] = 1;
  dfs_down(1, 0);
  dfs_result(1, 0);
  for (long long i = 1; i <= cd; i++) {
    result +=
        ((odd_down[i] * sz[i] % mod + odd_up[i] * (cd - sz[i])) % mod * v[i]) %
        mod;
    result %= mod;
    result -= ((even_down[i] * sz[i] % mod + even_up[i] * (cd - sz[i])) % mod *
               v[i]) %
              mod;
    result += mod * mod;
    result %= mod;
  }
  cout << result << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  inp();
  do_it();
}

```