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 = 2e5 + 5;
const long long mod = 1e9 + 7;
long long giv[maxn];
int n, nm;
long long w[2];
long long res;
struct node {
  int to, next;
} edg[maxn * 2];
int beg[maxn];
bool col[maxn];
long long num[maxn][2];
void ran(int x, bool y, int fa) {
  col[x] = y;
  w[y]++;
  for (int i = beg[x]; i != -1; i = edg[i].next) {
    if (edg[i].to == fa) continue;
    ran(edg[i].to, !y, x);
  }
}
void add(int a, int b) {
  edg[++nm].to = b;
  edg[nm].next = beg[a];
  beg[a] = nm;
}
void play(int x, int fa) {
  int i, j, a, b;
  for (i = beg[x]; i != -1; i = edg[i].next) {
    if (edg[i].to == fa) continue;
    play(edg[i].to, x);
    num[x][0] += num[edg[i].to][0];
    num[x][1] += num[edg[i].to][1];
    res += num[edg[i].to][col[x]] * (w[col[x]] - num[edg[i].to][col[x]]) % mod *
           giv[x] % mod;
    res -= num[edg[i].to][!col[x]] * (w[!col[x]] - num[edg[i].to][!col[x]]) %
           mod * giv[x] % mod;
    res = (res % mod + mod) % mod;
  }
  num[x][col[x]]++;
  res += num[x][col[x]] * (w[col[x]] - num[x][col[x]]) % mod * giv[x] % mod;
  res -= num[x][!col[x]] * (w[!col[x]] - num[x][!col[x]]) % mod * giv[x] % mod;
  res = (res % mod + mod) % mod;
}
int main() {
  memset(beg, -1, sizeof(beg));
  int i, j, a, b;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%lld", &giv[i]);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    add(a, b);
    add(b, a);
  }
  ran(1, 0, 0);
  for (i = 1; i <= n; i++) {
    res += giv[i] * w[col[i]];
    res %= mod;
  }
  play(1, 0);
  printf("%lld", res);
  return 0;
}
