#include <bits/stdc++.h>
using namespace std;
const long long int INF = 2000000000000000000;
long long int MOD(long long int a, long long int b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
long long int max3(long long int a, long long int b, long long int c) {
  return max(c, max(a, b));
}
long long int min3(long long int a, long long int b, long long int c) {
  return min(a, min(b, c));
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long int logg(long long int a) {
  long long int x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int factorial(long long int n) {
  if (n == 1 || n == 0) {
    return 1;
  } else {
    return factorial(n - 1);
  }
}
vector<long long int> v;
vector<vector<int>> adj;
vector<long long int> ans;
vector<long long int> visited;
vector<long long int> r;
long long int ans1 = 0;
long long int dfsu(int i, int j) {
  visited[i] = 1;
  long long int size = 1;
  for (int k = 0; k < adj[i].size(); k++) {
    if (visited[adj[i][k]] != 1) {
      size = size + dfsu(adj[i][k], j + 1);
    }
  }
  r[i] = size;
  return size;
}
void solve1() { dfsu(1, 1); }
void dfs(int i, int level) {
  visited[i] = 1;
  if (level % 2 == 1) {
    ans1 = ans1 + v[i] * r[i];
  } else {
    ans1 = ans1 - r[i] * v[i];
  }
  if (ans1 < 0) {
    ans1 = ans1 + (0 - ans1) / 1000000007 * 1000000007;
    ans1 = ans1 + 1000000007;
  }
  ans1 = ans1 % 1000000007;
  for (int k = 0; k < adj[i].size(); k++) {
    if (visited[adj[i][k]] != 1) {
      dfs(adj[i][k], level + 1);
    }
  }
}
void solve() { dfs(1, 1); }
void dfsu2(int i, int par, int n) {
  visited[i] = 1;
  ans[i] =
      -(ans[par] - n * v[par] + r[i] * v[i]) - (n - r[i]) * v[par] + n * v[i];
  if (ans[i] < 0) {
    ans[i] = ans[i] + (0 - ans[i]) / 1000000007 * 1000000007;
    ans[i] = ans[i] + 1000000007;
  }
  ans[i] = ans[i] % 1000000007;
  for (int k = 0; k < adj[i].size(); k++) {
    if (visited[adj[i][k]] != 1) {
      dfsu2(adj[i][k], i, n);
    }
  }
}
void solve2(int n) {
  visited[1] = 1;
  for (int j = 0; j < adj[1].size(); j++) {
    if (visited[adj[1][j]] != 1) {
      dfsu2(adj[1][j], 1, n);
    }
  }
}
int main() {
  int n;
  cin >> n;
  v.resize(n + 1, 0);
  adj.resize(n + 1);
  ans.resize(n + 1, 0);
  visited.resize(n + 1, 0);
  r.resize(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  solve1();
  for (int i = 0; i <= n; i++) {
    visited[i] = 0;
  }
  solve();
  ans[1] = ans1;
  for (int i = 0; i <= n; i++) {
    visited[i] = 0;
  }
  long long int ansall = 0;
  solve2(n);
  for (int i = 1; i <= n; i++) {
    ansall = ans[i] + ansall;
    if (ansall < 0) {
      ansall = ansall + (0 - ansall) / 1000000007 * 1000000007;
      ansall = ansall + 1000000007;
    }
    ansall = ansall % 1000000007;
  }
  ansall = ansall % 1000000007;
  cout << ansall % (1000000007);
}

### 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 INF = 2000000000000000000;
long long int MOD(long long int a, long long int b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
long long int max3(long long int a, long long int b, long long int c) {
  return max(c, max(a, b));
}
long long int min3(long long int a, long long int b, long long int c) {
  return min(a, min(b, c));
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long int logg(long long int a) {
  long long int x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int factorial(long long int n) {
  if (n == 1 || n == 0) {
    return 1;
  } else {
    return factorial(n - 1);
  }
}
vector<long long int> v;
vector<vector<int>> adj;
vector<long long int> ans;
vector<long long int> visited;
vector<long long int> r;
long long int ans1 = 0;
long long int dfsu(int i, int j) {
  visited[i] = 1;
  long long int size = 1;
  for (int k = 0; k < adj[i].size(); k++) {
    if (visited[adj[i][k]] != 1) {
      size = size + dfsu(adj[i][k], j + 1);
    }
  }
  r[i] = size;
  return size;
}
void solve1() { dfsu(1, 1); }
void dfs(int i, int level) {
  visited[i] = 1;
  if (level % 2 == 1) {
    ans1 = ans1 + v[i] * r[i];
  } else {
    ans1 = ans1 - r[i] * v[i];
  }
  if (ans1 < 0) {
    ans1 = ans1 + (0 - ans1) / 1000000007 * 1000000007;
    ans1 = ans1 + 1000000007;
  }
  ans1 = ans1 % 1000000007;
  for (int k = 0; k < adj[i].size(); k++) {
    if (visited[adj[i][k]] != 1) {
      dfs(adj[i][k], level + 1);
    }
  }
}
void solve() { dfs(1, 1); }
void dfsu2(int i, int par, int n) {
  visited[i] = 1;
  ans[i] =
      -(ans[par] - n * v[par] + r[i] * v[i]) - (n - r[i]) * v[par] + n * v[i];
  if (ans[i] < 0) {
    ans[i] = ans[i] + (0 - ans[i]) / 1000000007 * 1000000007;
    ans[i] = ans[i] + 1000000007;
  }
  ans[i] = ans[i] % 1000000007;
  for (int k = 0; k < adj[i].size(); k++) {
    if (visited[adj[i][k]] != 1) {
      dfsu2(adj[i][k], i, n);
    }
  }
}
void solve2(int n) {
  visited[1] = 1;
  for (int j = 0; j < adj[1].size(); j++) {
    if (visited[adj[1][j]] != 1) {
      dfsu2(adj[1][j], 1, n);
    }
  }
}
int main() {
  int n;
  cin >> n;
  v.resize(n + 1, 0);
  adj.resize(n + 1);
  ans.resize(n + 1, 0);
  visited.resize(n + 1, 0);
  r.resize(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  solve1();
  for (int i = 0; i <= n; i++) {
    visited[i] = 0;
  }
  solve();
  ans[1] = ans1;
  for (int i = 0; i <= n; i++) {
    visited[i] = 0;
  }
  long long int ansall = 0;
  solve2(n);
  for (int i = 1; i <= n; i++) {
    ansall = ans[i] + ansall;
    if (ansall < 0) {
      ansall = ansall + (0 - ansall) / 1000000007 * 1000000007;
      ansall = ansall + 1000000007;
    }
    ansall = ansall % 1000000007;
  }
  ansall = ansall % 1000000007;
  cout << ansall % (1000000007);
}

```