#include <bits/stdc++.h>
using namespace std;
const long long int modulo = 1e9 + 7;
void add(long long int &a, long long int b) { a = (a + b) % modulo; }
const int limit = 1000000;
int n;
vector<int> g[limit];
long long int value[limit];
int father[limit];
int numnodes[limit];
int numnodes0[limit];
int numnodes1[limit];
vector<int> lista[limit];
vector<int> lista0[limit];
vector<int> lista1[limit];
void precompute(int u, int p) {
  father[u] = p;
  numnodes[u] = 1;
  numnodes0[u] = 1;
  vector<int> &ar = g[u];
  vector<int> nextar;
  for (int i = 0; i < int(ar.size()); i++) {
    int v = ar[i];
    if (v != p) {
      precompute(v, u);
      numnodes[u] += numnodes[v];
      numnodes0[u] += numnodes1[v];
      numnodes1[u] += numnodes0[v];
      lista[u].push_back(numnodes[v]);
      lista0[u].push_back(numnodes1[v]);
      lista1[u].push_back(numnodes0[v]);
      nextar.push_back(v);
    }
  }
  ar = nextar;
}
long long int sol;
void compute(int u) {
  if (u != 1) {
    int p = father[u];
    lista[u].push_back(numnodes[p] - numnodes[u]);
    lista0[u].push_back(numnodes1[p] - numnodes0[u]);
    lista1[u].push_back(numnodes0[p] - numnodes1[u]);
    numnodes[u] = numnodes[p];
    numnodes0[u] = numnodes1[p];
    numnodes1[u] = numnodes0[p];
  }
  long long int val = value[u];
  vector<int> &lis = lista[u];
  vector<int> &lis0 = lista0[u];
  vector<int> &lis1 = lista1[u];
  add(sol, val * n);
  for (int i = 0; i < int(lis.size()); i++) {
    long long int num = lis[i];
    long long int num0 = lis0[i];
    long long int num1 = lis1[i];
    add(sol, val * num0 % modulo * (n - num));
    add(sol, -val * num1 % modulo * (n - num));
  }
  vector<int> &ar = g[u];
  for (int i = 0; i < int(ar.size()); i++) compute(ar[i]);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> value[i];
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  precompute(1, 0);
  compute(1);
  sol = (sol % modulo + modulo) % modulo;
  cout << sol << endl;
}

### Prompt

Your challenge is to write 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 long long int modulo = 1e9 + 7;
void add(long long int &a, long long int b) { a = (a + b) % modulo; }
const int limit = 1000000;
int n;
vector<int> g[limit];
long long int value[limit];
int father[limit];
int numnodes[limit];
int numnodes0[limit];
int numnodes1[limit];
vector<int> lista[limit];
vector<int> lista0[limit];
vector<int> lista1[limit];
void precompute(int u, int p) {
  father[u] = p;
  numnodes[u] = 1;
  numnodes0[u] = 1;
  vector<int> &ar = g[u];
  vector<int> nextar;
  for (int i = 0; i < int(ar.size()); i++) {
    int v = ar[i];
    if (v != p) {
      precompute(v, u);
      numnodes[u] += numnodes[v];
      numnodes0[u] += numnodes1[v];
      numnodes1[u] += numnodes0[v];
      lista[u].push_back(numnodes[v]);
      lista0[u].push_back(numnodes1[v]);
      lista1[u].push_back(numnodes0[v]);
      nextar.push_back(v);
    }
  }
  ar = nextar;
}
long long int sol;
void compute(int u) {
  if (u != 1) {
    int p = father[u];
    lista[u].push_back(numnodes[p] - numnodes[u]);
    lista0[u].push_back(numnodes1[p] - numnodes0[u]);
    lista1[u].push_back(numnodes0[p] - numnodes1[u]);
    numnodes[u] = numnodes[p];
    numnodes0[u] = numnodes1[p];
    numnodes1[u] = numnodes0[p];
  }
  long long int val = value[u];
  vector<int> &lis = lista[u];
  vector<int> &lis0 = lista0[u];
  vector<int> &lis1 = lista1[u];
  add(sol, val * n);
  for (int i = 0; i < int(lis.size()); i++) {
    long long int num = lis[i];
    long long int num0 = lis0[i];
    long long int num1 = lis1[i];
    add(sol, val * num0 % modulo * (n - num));
    add(sol, -val * num1 % modulo * (n - num));
  }
  vector<int> &ar = g[u];
  for (int i = 0; i < int(ar.size()); i++) compute(ar[i]);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> value[i];
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  precompute(1, 0);
  compute(1);
  sol = (sol % modulo + modulo) % modulo;
  cout << sol << endl;
}

```