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 maxn = 200003, mod = 1000000007;
int Plus(int x, int y) { return (x += y) >= mod ? x - mod : x; }
void PlusEqual(int &x, int y) {
  if ((x += y) >= mod) x -= mod;
}
int Minus(int x, int y) { return Plus(x, mod - y); }
int mul(long long x, int y) { return x * y % mod; }
vector<int> g[maxn];
int n, a[maxn], dp[maxn][2], cnt[maxn][2], ans;
void dfs(int u, int last) {
  int val[2] = {a[u], Minus(0, a[u])};
  for (int v : g[u]) {
    if (v == last) continue;
    dfs(v, u);
    for (int j = 0; j <= 1; j++) {
      PlusEqual(ans,
                Plus(mul(dp[u][j], cnt[v][!j]), mul(dp[v][!j], cnt[u][j])));
      PlusEqual(dp[u][j], Plus(dp[v][!j], mul(cnt[v][!j], val[j])));
      PlusEqual(cnt[u][j], cnt[v][!j]);
    }
  }
  PlusEqual(ans, dp[u][0]);
  PlusEqual(dp[u][0], val[0]);
  PlusEqual(cnt[u][0], 1);
}
int main() {
  scanf("%d", &n);
  int sum = 0;
  for (int i = 1; i <= n; i++)
    scanf("%d", a + i), a[i] = (a[i] + ((long long)mod << 2)) % mod,
                        PlusEqual(sum, a[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v), g[v].push_back(u);
  }
  dfs(1, 0);
  printf("%d\n", Plus(mul(ans, 2), sum));
  return 0;
}
