#include <bits/stdc++.h>
using namespace std;
vector<int> krawedzie[200007];
long long waga[200007];
int ojciec[200007];
int warstwa[200007];
long long parz = 0;
long long nparz = 0;
long long wynik = 0;
void dfs1(int u) {
  if (warstwa[u] % 2 == 0) {
    parz++;
  } else {
    nparz++;
  }
  for (int i = 0; i < krawedzie[u].size(); i++) {
    int v = krawedzie[u][i];
    if (v != ojciec[u]) {
      warstwa[v] = warstwa[u] + 1;
      ojciec[v] = u;
      dfs1(v);
    }
  }
}
pair<long long, long long> dfs2(int u) {
  long long p = 0;
  long long np = 0;
  long long mn = 1;
  if (warstwa[u] % 2 == 1) {
    mn = -1;
  }
  for (int i = 0; i < krawedzie[u].size(); i++) {
    int v = krawedzie[u][i];
    if (v != ojciec[u]) {
      pair<long long, long long> tmp = dfs2(v);
      wynik +=
          (mn * (((p + np) * tmp.first) % 1000000007) * waga[u]) % 1000000007;
      wynik = wynik % 1000000007;
      wynik -=
          (mn * (((p + np) * tmp.second) % 1000000007) * waga[u]) % 1000000007;
      wynik = wynik % 1000000007;
      wynik += (mn * (((tmp.first + tmp.second) * p) % 1000000007) * waga[u]) %
               1000000007;
      wynik = wynik % 1000000007;
      wynik -= (mn * (((tmp.first + tmp.second) * np) % 1000000007) * waga[u]) %
               1000000007;
      wynik = wynik % 1000000007;
      p += tmp.first;
      np += tmp.second;
    }
  }
  wynik += (mn * ((p * (parz + nparz - p - np)) % 1000000007) * waga[u]) %
           1000000007;
  wynik = wynik % 1000000007;
  wynik -= (mn * ((np * (parz + nparz - p - np)) % 1000000007) * waga[u]) %
           1000000007;
  wynik = wynik % 1000000007;
  wynik += (mn * (((parz - p) * (p + np)) % 1000000007) * waga[u]) % 1000000007;
  wynik = wynik % 1000000007;
  wynik -=
      (mn * (((nparz - np) * (p + np)) % 1000000007) * waga[u]) % 1000000007;
  wynik = wynik % 1000000007;
  wynik += (waga[u] * (parz + nparz - p - np)) % 1000000007;
  wynik = wynik % 1000000007;
  if (warstwa[u] % 2 == 0) {
    p++;
  } else {
    np++;
  }
  wynik += (((mn * (parz - p)) % 1000000007) * waga[u]) % 1000000007;
  wynik -= (((mn * (nparz - np)) % 1000000007) * waga[u]) % 1000000007;
  wynik = wynik % 1000000007;
  return make_pair(p, np);
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> waga[i];
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    krawedzie[a].push_back(b);
    krawedzie[b].push_back(a);
  }
  dfs1(0);
  dfs2(0);
  if (wynik < 0) {
    wynik += 1000000007;
  }
  cout << wynik;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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<int> krawedzie[200007];
long long waga[200007];
int ojciec[200007];
int warstwa[200007];
long long parz = 0;
long long nparz = 0;
long long wynik = 0;
void dfs1(int u) {
  if (warstwa[u] % 2 == 0) {
    parz++;
  } else {
    nparz++;
  }
  for (int i = 0; i < krawedzie[u].size(); i++) {
    int v = krawedzie[u][i];
    if (v != ojciec[u]) {
      warstwa[v] = warstwa[u] + 1;
      ojciec[v] = u;
      dfs1(v);
    }
  }
}
pair<long long, long long> dfs2(int u) {
  long long p = 0;
  long long np = 0;
  long long mn = 1;
  if (warstwa[u] % 2 == 1) {
    mn = -1;
  }
  for (int i = 0; i < krawedzie[u].size(); i++) {
    int v = krawedzie[u][i];
    if (v != ojciec[u]) {
      pair<long long, long long> tmp = dfs2(v);
      wynik +=
          (mn * (((p + np) * tmp.first) % 1000000007) * waga[u]) % 1000000007;
      wynik = wynik % 1000000007;
      wynik -=
          (mn * (((p + np) * tmp.second) % 1000000007) * waga[u]) % 1000000007;
      wynik = wynik % 1000000007;
      wynik += (mn * (((tmp.first + tmp.second) * p) % 1000000007) * waga[u]) %
               1000000007;
      wynik = wynik % 1000000007;
      wynik -= (mn * (((tmp.first + tmp.second) * np) % 1000000007) * waga[u]) %
               1000000007;
      wynik = wynik % 1000000007;
      p += tmp.first;
      np += tmp.second;
    }
  }
  wynik += (mn * ((p * (parz + nparz - p - np)) % 1000000007) * waga[u]) %
           1000000007;
  wynik = wynik % 1000000007;
  wynik -= (mn * ((np * (parz + nparz - p - np)) % 1000000007) * waga[u]) %
           1000000007;
  wynik = wynik % 1000000007;
  wynik += (mn * (((parz - p) * (p + np)) % 1000000007) * waga[u]) % 1000000007;
  wynik = wynik % 1000000007;
  wynik -=
      (mn * (((nparz - np) * (p + np)) % 1000000007) * waga[u]) % 1000000007;
  wynik = wynik % 1000000007;
  wynik += (waga[u] * (parz + nparz - p - np)) % 1000000007;
  wynik = wynik % 1000000007;
  if (warstwa[u] % 2 == 0) {
    p++;
  } else {
    np++;
  }
  wynik += (((mn * (parz - p)) % 1000000007) * waga[u]) % 1000000007;
  wynik -= (((mn * (nparz - np)) % 1000000007) * waga[u]) % 1000000007;
  wynik = wynik % 1000000007;
  return make_pair(p, np);
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> waga[i];
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    krawedzie[a].push_back(b);
    krawedzie[b].push_back(a);
  }
  dfs1(0);
  dfs2(0);
  if (wynik < 0) {
    wynik += 1000000007;
  }
  cout << wynik;
  return 0;
}

```