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 long long MOD = 1000 * 1000 * 1000 + 7;
const long long MAXN = 2e5 + 7;
long long a[MAXN];
vector<long long> g[MAXN];
long long mod(long long n) { return (n % MOD + MOD) % MOD; }
long long t1[MAXN], t2[MAXN], cnt[MAXN];
void calc(long long u, long long parent, long long h) {
  if (h % 2)
    ++t1[u];
  else
    ++t2[u];
  cnt[u] = 1;
  for (long long v : g[u]) {
    if (v == parent) continue;
    calc(v, u, h + 1);
    t1[u] += t1[v];
    t2[u] += t2[v];
    cnt[u] += cnt[v];
  }
}
long long n;
long long ans = 0;
void jfs(long long u, long long parent, long long h) {
  for (long long v : g[u]) {
    if (v == parent) continue;
    jfs(v, u, h + 1);
  }
  ans += a[u] * n;
  ans = mod(ans);
  long long sum1 = t1[0];
  long long sum2 = t2[0];
  for (long long v : g[u]) {
    long long n1, n2, sz;
    if (v == parent) {
      n1 = sum1 - t1[u];
      n2 = sum2 - t2[u];
      sz = cnt[u];
    } else {
      n1 = t1[v];
      n2 = t2[v];
      sz = n - cnt[v];
    }
    if (h % 2) {
      ans += n1 * sz * a[u];
      ans = mod(ans);
      ans -= n2 * sz * a[u];
      ans = mod(ans);
    } else {
      ans -= n1 * sz * a[u];
      ans = mod(ans);
      ans += n2 * sz * a[u];
      ans = mod(ans);
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = 0; i < n - 1; ++i) {
    long long u, v;
    cin >> u >> v;
    g[u - 1].push_back(v - 1);
    g[v - 1].push_back(u - 1);
  }
  calc(0, 0, 0);
  jfs(0, 0, 0);
  cout << ans << '\n';
  return 0;
}
