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 int val[200003], eve[200003], odd[200003], ans, cnte[200003],
    cnto[200003];
vector<int> v[200003];
void dfs(int node, int par) {
  int i;
  long long int e = val[node], o = 0, ce = 1, co = 0;
  ans = (ans + e) % 1000000007;
  for (i = 0; i < v[node].size(); i++) {
    int ch = v[node][i];
    if (ch == par) continue;
    dfs(ch, node);
    ans = (ans + 2 * (e * cnto[ch] + odd[ch] * ce)) % 1000000007;
    ans = (ans + 2 * (o * cnte[ch] + eve[ch] * co)) % 1000000007;
    e = (e + odd[ch] + cnto[ch] * val[node]) % 1000000007;
    ce = (ce + cnto[ch]) % 1000000007;
    o = (o + eve[ch] - (cnte[ch] * val[node]) % 1000000007 + 1000000007) %
        1000000007;
    co = (co + cnte[ch]) % 1000000007;
  }
  eve[node] = e;
  odd[node] = o;
  cnte[node] = ce;
  cnto[node] = co;
}
int main() {
  int i, n;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%lld", &val[i]);
    val[i] = (1000000007 + val[i] % 1000000007) % 1000000007;
  }
  for (i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(1, -1);
  printf("%lld\n", ans);
  return 0;
}
