#include <bits/stdc++.h>
using namespace std;
const long long MAXn = 2e5 + 10, MOD = 1e9 + 7;
long long n, odd[MAXn], even[MAXn], h[2], ar[MAXn], ans;
vector<int> g[MAXn];
void dfs(int a, int par, int dis) {
  h[dis % 2]++;
  for (int i = 0; i < g[a].size(); i++)
    if (g[a][i] != par) {
      dfs(g[a][i], a, dis + 1);
      even[a] += odd[g[a][i]];
      odd[a] += even[g[a][i]];
    }
  for (int i = 0; i < g[a].size(); i++)
    if (g[a][i] != par) {
      long long odd1 = odd[a] - even[g[a][i]], even1 = even[a] - odd[g[a][i]];
      ans += (((((even1 - odd1 + MOD) % MOD) * (odd[g[a][i]] + even[g[a][i]])) %
               MOD) *
              ar[a]) %
             MOD;
      ans %= MOD;
    }
  ans += ((odd[a] + even[a]) * ar[a]) % MOD;
  ans %= MOD;
  ans += (((even[a] - odd[a] + MOD) % MOD) * ar[a]) % MOD;
  ans %= MOD;
  ans += ar[a];
  ans %= MOD;
  even[a]++;
}
void dfs2(int a, int par, int dis) {
  ans += (((((even[a] - odd[a] + MOD) % MOD) * (n - even[a] - odd[a])) % MOD) *
          ar[a]) %
         MOD;
  ans %= MOD;
  long long odd1 = h[(dis + 1) % 2] - odd[a], even1 = h[(dis) % 2] - even[a];
  ans += (((((even1 - odd1 + MOD) % MOD) * (even[a] + odd[a])) % MOD) * ar[a]) %
         MOD;
  ans %= MOD;
  for (int i = 0; i < g[a].size(); i++)
    if (g[a][i] != par) dfs2(g[a][i], a, dis + 1);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    ar[i] += MOD;
    ar[i] %= MOD;
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, -1, 0);
  dfs2(0, -1, 0);
  cout << ans;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 long long MAXn = 2e5 + 10, MOD = 1e9 + 7;
long long n, odd[MAXn], even[MAXn], h[2], ar[MAXn], ans;
vector<int> g[MAXn];
void dfs(int a, int par, int dis) {
  h[dis % 2]++;
  for (int i = 0; i < g[a].size(); i++)
    if (g[a][i] != par) {
      dfs(g[a][i], a, dis + 1);
      even[a] += odd[g[a][i]];
      odd[a] += even[g[a][i]];
    }
  for (int i = 0; i < g[a].size(); i++)
    if (g[a][i] != par) {
      long long odd1 = odd[a] - even[g[a][i]], even1 = even[a] - odd[g[a][i]];
      ans += (((((even1 - odd1 + MOD) % MOD) * (odd[g[a][i]] + even[g[a][i]])) %
               MOD) *
              ar[a]) %
             MOD;
      ans %= MOD;
    }
  ans += ((odd[a] + even[a]) * ar[a]) % MOD;
  ans %= MOD;
  ans += (((even[a] - odd[a] + MOD) % MOD) * ar[a]) % MOD;
  ans %= MOD;
  ans += ar[a];
  ans %= MOD;
  even[a]++;
}
void dfs2(int a, int par, int dis) {
  ans += (((((even[a] - odd[a] + MOD) % MOD) * (n - even[a] - odd[a])) % MOD) *
          ar[a]) %
         MOD;
  ans %= MOD;
  long long odd1 = h[(dis + 1) % 2] - odd[a], even1 = h[(dis) % 2] - even[a];
  ans += (((((even1 - odd1 + MOD) % MOD) * (even[a] + odd[a])) % MOD) * ar[a]) %
         MOD;
  ans %= MOD;
  for (int i = 0; i < g[a].size(); i++)
    if (g[a][i] != par) dfs2(g[a][i], a, dis + 1);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    ar[i] += MOD;
    ar[i] %= MOD;
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, -1, 0);
  dfs2(0, -1, 0);
  cout << ans;
}

```