#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > g;
vector<long long> dp[5];
vector<vector<long long> > tr[5];
long long val[200200];
const long long M = 1e9 + 7;
void calc(int u, int p) {
  long long tot = 0;
  long long pv[5] = {0, 0, 0, 0, 0};
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    calc(v, u);
    tot += dp[2][v];
    for (int j = 0; j < 5; j++) {
      pv[j] += dp[j][v];
    }
  }
  for (int i = 0; i < 5; i++) {
    tr[i][u].assign(g[u].size(), 0);
  }
  long long rs[5] = {0, 0, 0, 0, 0};
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    for (int j = 0; j < 5; j++) {
      tr[j][u][i] += rs[j];
    }
    for (int j = 0; j < 5; j++) {
      rs[j] += dp[j][v];
    }
  }
  for (int i = 0; i < 5; i++) {
    rs[i] = 0;
  }
  for (int i = g[u].size() - 1; i >= 0; i--) {
    int v = g[u][i];
    if (v == p) continue;
    for (int j = 0; j < 5; j++) {
      tr[j][u][i] += rs[j];
      tr[j][u][i] %= M;
    }
    for (int j = 0; j < 5; j++) {
      rs[j] += dp[j][v];
    }
  }
  for (int i = 0; i < 5; i++) {
    pv[i] %= M;
  }
  tot += pv[1] + pv[0] + val[u] * (pv[4] - pv[3]);
  tot += val[u] * (pv[3] + pv[4]) - pv[0] + pv[1];
  tot %= M;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    tot += (val[u] * (dp[4][v] - dp[3][v])) % M * (tr[4][u][i] + tr[3][u][i]);
    tot += (-tr[0][u][i] + tr[1][u][i]) * (-dp[3][v] + dp[4][v]);
    tot += (dp[1][v] + dp[0][v]) * (tr[4][u][i] + tr[3][u][i]);
    tot %= M;
  }
  dp[3][u] = pv[4] + 1;
  dp[4][u] = pv[3];
  dp[0][u] = pv[1] + val[u] * dp[3][u];
  dp[1][u] = pv[0] - val[u] * dp[4][u];
  dp[2][u] = tot + val[u];
  for (int i = 0; i < 5; i++) {
    dp[i][u] %= M;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  g.assign(n, vector<long long>());
  for (int i = 0; i < 5; i++) {
    dp[i].assign(n, 0);
    tr[i].assign(n, vector<long long>());
  }
  for (int i = 0; i < n; i++) {
    cin >> val[i];
  }
  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);
  }
  calc(0, -1);
  long long res = dp[2][0];
  res = ((res % M) + M) % M;
  cout << res << '\n';
}

### Prompt

Create a solution in Cpp for 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;
vector<vector<long long> > g;
vector<long long> dp[5];
vector<vector<long long> > tr[5];
long long val[200200];
const long long M = 1e9 + 7;
void calc(int u, int p) {
  long long tot = 0;
  long long pv[5] = {0, 0, 0, 0, 0};
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    calc(v, u);
    tot += dp[2][v];
    for (int j = 0; j < 5; j++) {
      pv[j] += dp[j][v];
    }
  }
  for (int i = 0; i < 5; i++) {
    tr[i][u].assign(g[u].size(), 0);
  }
  long long rs[5] = {0, 0, 0, 0, 0};
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    for (int j = 0; j < 5; j++) {
      tr[j][u][i] += rs[j];
    }
    for (int j = 0; j < 5; j++) {
      rs[j] += dp[j][v];
    }
  }
  for (int i = 0; i < 5; i++) {
    rs[i] = 0;
  }
  for (int i = g[u].size() - 1; i >= 0; i--) {
    int v = g[u][i];
    if (v == p) continue;
    for (int j = 0; j < 5; j++) {
      tr[j][u][i] += rs[j];
      tr[j][u][i] %= M;
    }
    for (int j = 0; j < 5; j++) {
      rs[j] += dp[j][v];
    }
  }
  for (int i = 0; i < 5; i++) {
    pv[i] %= M;
  }
  tot += pv[1] + pv[0] + val[u] * (pv[4] - pv[3]);
  tot += val[u] * (pv[3] + pv[4]) - pv[0] + pv[1];
  tot %= M;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    tot += (val[u] * (dp[4][v] - dp[3][v])) % M * (tr[4][u][i] + tr[3][u][i]);
    tot += (-tr[0][u][i] + tr[1][u][i]) * (-dp[3][v] + dp[4][v]);
    tot += (dp[1][v] + dp[0][v]) * (tr[4][u][i] + tr[3][u][i]);
    tot %= M;
  }
  dp[3][u] = pv[4] + 1;
  dp[4][u] = pv[3];
  dp[0][u] = pv[1] + val[u] * dp[3][u];
  dp[1][u] = pv[0] - val[u] * dp[4][u];
  dp[2][u] = tot + val[u];
  for (int i = 0; i < 5; i++) {
    dp[i][u] %= M;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  g.assign(n, vector<long long>());
  for (int i = 0; i < 5; i++) {
    dp[i].assign(n, 0);
    tr[i].assign(n, vector<long long>());
  }
  for (int i = 0; i < n; i++) {
    cin >> val[i];
  }
  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);
  }
  calc(0, -1);
  long long res = dp[2][0];
  res = ((res % M) + M) % M;
  cout << res << '\n';
}

```