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. 
#include <bits/stdc++.h>
using namespace std;
long long V[1000000];
long long mod = 1000000007;
long long sum[1000000][2];
long long num[1000000][2];
long long cur[1000000][2];
vector<int> adj[1000000];
long long ret = 0;
void dfs(int v, int p) {
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (u == p) continue;
    dfs(u, v);
    cur[v][0] += sum[u][0];
    cur[v][1] += sum[u][1];
    cur[v][0] %= mod;
    cur[v][1] %= mod;
    sum[v][0] += (sum[u][1] - (num[u][1] * V[v]) % mod + mod + mod) % mod;
    sum[v][0] %= mod;
    num[v][0] += num[u][0];
    num[v][1] += num[u][1];
    sum[v][1] += (sum[u][0] + (num[u][0] * V[v]) % mod) % mod;
    sum[v][1] %= mod;
  }
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (u == p) continue;
    for (int j = 0; j < 2; j++) {
      if (j == 0) V[v] = -V[v];
      ret +=
          ((-V[v] * ((num[u][j] * (num[v][j] - num[u][j])) % mod) + mod) % mod);
      ret %= mod;
      ret += (num[u][j] * ((cur[v][j] - sum[u][j] + mod) % mod)) % mod;
      ret %= mod;
      ret += ((num[v][j] - num[u][j]) % mod) * (sum[u][j]) % mod;
      ret %= mod;
      if (j == 0) V[v] = -V[v];
    }
  }
  swap(num[v][0], num[v][1]);
  num[v][1]++;
  ret += (sum[v][1] * 2) % mod;
  sum[v][1] += V[v] + mod;
  sum[v][1] %= mod;
  ret %= mod;
  ret += (V[v] + mod);
  ret %= mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> V[i], V[i] += mod, V[i] %= mod;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, -1);
  cout << ret << endl;
}
