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 kMod = 1e9 + 7;
const int kNmax = 2e5 + 10;
int n, v[kNmax], sub[kNmax], up[kNmax][2], down[kNmax][2], res, res_even;
vector<int> g[kNmax];
void dfs(int node, int father = 0) {
  sub[node] = 1;
  down[node][0] = 1;
  int ways = (1LL * n * n) % kMod;
  for (int vec : g[node]) {
    if (vec == father) continue;
    dfs(vec, node);
    ways = (ways - 1LL * sub[vec] * sub[vec]) % kMod;
    if (ways < 0) ways += kMod;
    sub[node] += sub[vec];
    down[node][0] += down[vec][1];
    down[node][1] += down[vec][0];
  }
  ways = (ways - 1LL * (n - sub[node]) * (n - sub[node])) % kMod;
  if (ways < 0) ways += kMod;
  res = (res + 1LL * ways * v[node]) % kMod;
  if (res < 0) res += kMod;
}
void dfs_bad(int node, int father = 0) {
  int s0 = 0, s1 = 0;
  int ways = (1LL * down[node][1] * (n - sub[node])) % kMod;
  ways = (ways + 1LL * up[node][1] * sub[node]) % kMod;
  for (int vec : g[node]) {
    if (vec == father) continue;
    ways = (ways + 1LL * down[vec][0] * (sub[node] - sub[vec])) % kMod;
    s0 += down[vec][0];
    s1 += down[vec][1];
  }
  res_even = (res_even + 1LL * ways * v[node]) % kMod;
  if (res_even < 0) res_even += kMod;
  for (int vec : g[node]) {
    if (vec == father) continue;
    up[vec][0] = up[node][1] + s0 - down[vec][0];
    up[vec][1] = n - sub[vec] - up[vec][0];
    dfs_bad(vec, node);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> v[i];
  for (int i = 1, a, b; i < n; i++) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1);
  up[1][0] = 1;
  dfs_bad(1);
  res = (res - 2LL * res_even) % kMod;
  if (res < 0) res += kMod;
  cout << res;
}
