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 int N = 2e5 + 5, mod = 1e9 + 7;
int n, a[N], siz[N];
long long fd[N], gd[N], dp[N], fu[N], gu[N], ans;
vector<int> G[N];
void dfs(int x, int fa) {
  siz[x] = fd[x] = 1;
  for (auto it : G[x])
    if (it != fa) {
      dfs(it, x);
      siz[x] += siz[it];
      fd[x] += gd[it];
      gd[x] += fd[it];
    }
  ans = (ans + a[x] * (fd[x] - gd[x]) % mod * (n - siz[x] + 1)) % mod;
  for (auto it : G[x])
    if (it ^ fa)
      ans = (ans +
             a[x] * (gd[it] - fd[it]) % mod * (siz[x] - siz[it] - 1) % mod) %
            mod;
  ans = (ans + 1ll * a[x] * (siz[x] - 1)) % mod;
}
void dfs2(int x, int fa) {
  if (fa) {
    fu[x] = gu[fa] + gd[fa] - fd[x];
    gu[x] = fu[fa] + fd[fa] - gd[x];
  }
  for (auto it : G[x])
    if (it != fa) dfs2(it, x);
  ans = (ans + a[x] * (fu[x] - gu[x]) % mod * siz[x] % mod) % mod;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1, x, y; i < n; i++)
    cin >> x >> y, G[x].push_back(y), G[y].push_back(x);
  dfs(1, 0);
  dfs2(1, 0);
  cout << (ans + mod) % mod;
  return 0;
}
