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;
inline int in() {
  int x, y;
  y = scanf("%d", &x);
  return x;
}
const int N = 200 * 1000 + 5;
const int P = 1e9 + 7;
int a[N];
int sub[N][2], h[N];
long long ans = 0;
vector<int> adj[N];
int sum(int x, int y) {
  int res = x + y;
  if (res < 0) return (res % P + P) % P;
  if (res >= P) return res % P;
  return res;
}
int mul(int x, int y) {
  long long res = (long long)x * y;
  if (res < 0) return (res % P + P) % P;
  if (res >= P) return res % P;
  return res;
}
void dfs(int v, int p = -1) {
  int r = h[v] % 2;
  sub[v][r]++;
  for (int u : adj[v]) {
    if (u == p) continue;
    h[u] = h[v] + 1;
    dfs(u, v);
    sub[v][0] = sum(sub[v][0], sub[u][0]);
    sub[v][1] = sum(sub[v][1], sub[u][1]);
  }
}
void dfs2(int v, int p = -1) {
  int r = h[v] % 2;
  long long same = 1, diff = 0;
  ans = sum(ans, a[v]);
  for (int u : adj[v]) {
    long long scur, dcur;
    if (u == p) {
      scur = sum(sub[0][r], -sub[v][r]);
      dcur = sum(sub[0][1 - r], -sub[v][1 - r]);
    } else {
      dfs2(u, v);
      scur = sub[u][r];
      dcur = sub[u][1 - r];
    }
    ans = sum(ans, mul(mul(mul(same, scur), 2), a[v]));
    same = sum(same, scur);
    ans = sum(ans, mul(mul(mul(diff, dcur), 2), -a[v]));
    diff = sum(diff, dcur);
  }
}
int main() {
  int n = in();
  for (int i = 0; i < n; i++) a[i] = in();
  for (int i = 0; i < n - 1; i++) {
    int u = in() - 1, v = in() - 1;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(0);
  dfs2(0);
  cout << ans << '\n';
  return 0;
}
