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;
const long long m = 1000000007;
const int maxn = 200010;
vector<int> g[maxn];
int val[maxn];
int dep[maxn];
long long odd[maxn], even[maxn];
long long ans;
int n;
void dfs(int u, int fa, int d) {
  dep[u] = d;
  for (int v : g[u]) {
    if (v == fa) continue;
    if (v != fa) dfs(v, u, d + 1);
    ans = ((ans +
            2 * (even[v] * odd[u] % m - odd[v] * even[u] % m + m) % m * val[u] %
                m +
            m) %
               m +
           m) %
          m;
    odd[u] += even[v];
    even[u] += odd[v];
  }
  odd[u]++;
}
int main() {
  ios::sync_with_stdio(false);
  while (cin >> n) {
    for (int i = 0; i < n; i++) g[i].clear();
    memset(odd, 0, sizeof odd);
    memset(even, 0, sizeof even);
    ans = 0;
    for (int i = 1; i <= n; i++) {
      int v;
      cin >> v;
      val[i] = v;
    }
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    dfs(1, -1, 1);
    for (int i = 1; i <= n; i++) {
      if (dep[i] % 2 == 0) {
        ans = ((ans + 2 *
                          (odd[i] * (even[1] - odd[i] + 1) % m -
                           even[i] * (odd[1] - even[i]) % m + m) %
                          m * val[i] % m) %
                   m +
               m) %
              m;
      } else {
        ans = ((ans + 2 *
                          (odd[i] * (odd[1] - odd[i] + 1) % m -
                           even[i] * (even[1] - even[i]) % m + m) %
                          m * val[i] % m) %
                   m +
               m) %
              m;
      }
      ans = ((ans - val[i]) % m + m) % m;
    }
    cout << ans << endl;
  }
  return 0;
}
