#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int sum(int a, int b) { return (a + b) % MOD; }
int prod(int a, int b) { return a * 1ll * b % MOD; }
struct data {
  long long ans;
  long long op, ep;
  long long os, es;
  data add_root_to_paths(long long rv) {
    return {ans, ep + 1, op, sum(es, prod(ep + 1, rv)),
            sum(os, MOD - prod(op, rv))};
  }
};
data solve(vector<vector<int> > &tree, vector<long long> &val, int loc,
           int par) {
  data res = {0, 0, 0, 0, 0};
  for (int ch : tree[loc]) {
    if (ch == par) continue;
    data sub = solve(tree, val, ch, loc);
    for (int i = 0; i < 2; i++) {
      res.ans = sum(res.ans, sum(sub.es, prod(val[loc], sub.ep)));
      res.ans = sum(res.ans, prod(sub.es, res.ep));
      res.ans = sum(res.ans, prod(val[loc], prod(sub.ep, res.ep)));
      res.ans = sum(res.ans, prod(res.es, sub.ep));
      res.ans = sum(res.ans, prod(sub.os, res.op));
      res.ans = sum(res.ans, MOD - prod(val[loc], prod(sub.op, res.op)));
      res.ans = sum(res.ans, prod(res.os, sub.op));
    }
    res.ans = sum(res.ans, sub.ans);
    res.op += sub.op;
    res.os = sum(res.os, sub.os);
    res.ep += sub.ep;
    res.es = sum(res.es, sub.es);
  }
  res.ans = sum(res.ans, val[loc]);
  return res.add_root_to_paths(val[loc]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  int N;
  cin >> N;
  vector<long long> val(N);
  for (int i = 0; i < N; i++) {
    cin >> val[i];
    val[i] %= MOD;
    if (val[i] < 0) val[i] += MOD;
  }
  vector<vector<int> > adj(N);
  for (int i = 0; i < N - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u - 1].push_back(v - 1);
    adj[v - 1].push_back(u - 1);
  }
  cout << solve(adj, val, 0, 0).ans << endl;
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int sum(int a, int b) { return (a + b) % MOD; }
int prod(int a, int b) { return a * 1ll * b % MOD; }
struct data {
  long long ans;
  long long op, ep;
  long long os, es;
  data add_root_to_paths(long long rv) {
    return {ans, ep + 1, op, sum(es, prod(ep + 1, rv)),
            sum(os, MOD - prod(op, rv))};
  }
};
data solve(vector<vector<int> > &tree, vector<long long> &val, int loc,
           int par) {
  data res = {0, 0, 0, 0, 0};
  for (int ch : tree[loc]) {
    if (ch == par) continue;
    data sub = solve(tree, val, ch, loc);
    for (int i = 0; i < 2; i++) {
      res.ans = sum(res.ans, sum(sub.es, prod(val[loc], sub.ep)));
      res.ans = sum(res.ans, prod(sub.es, res.ep));
      res.ans = sum(res.ans, prod(val[loc], prod(sub.ep, res.ep)));
      res.ans = sum(res.ans, prod(res.es, sub.ep));
      res.ans = sum(res.ans, prod(sub.os, res.op));
      res.ans = sum(res.ans, MOD - prod(val[loc], prod(sub.op, res.op)));
      res.ans = sum(res.ans, prod(res.os, sub.op));
    }
    res.ans = sum(res.ans, sub.ans);
    res.op += sub.op;
    res.os = sum(res.os, sub.os);
    res.ep += sub.ep;
    res.es = sum(res.es, sub.es);
  }
  res.ans = sum(res.ans, val[loc]);
  return res.add_root_to_paths(val[loc]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  int N;
  cin >> N;
  vector<long long> val(N);
  for (int i = 0; i < N; i++) {
    cin >> val[i];
    val[i] %= MOD;
    if (val[i] < 0) val[i] += MOD;
  }
  vector<vector<int> > adj(N);
  for (int i = 0; i < N - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u - 1].push_back(v - 1);
    adj[v - 1].push_back(u - 1);
  }
  cout << solve(adj, val, 0, 0).ans << endl;
  return 0;
}

```