#include <bits/stdc++.h>
using namespace std;
int n, w[200010], size[200010], ans, g[200010][2];
vector<int> a[200010];
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
int inc(int x, int y) {
  return (x + y >= 1000000007) ? (x + y - 1000000007) : (x + y);
}
int dec(int x, int y) { return (x - y < 0) ? (x - y + 1000000007) : (x - y); }
int mul(int x, int y) { return 1ll * x * y % 1000000007; }
void dfs(int pos, int dad, int dep) {
  size[pos] = 1, g[pos][dep & 1] = 1;
  for (unsigned i = 0, si = a[pos].size(); i < si; ++i) {
    int son = a[pos][i];
    if (son != dad) {
      dfs(son, pos, dep + 1);
      size[pos] += size[son];
      g[pos][0] += g[son][0], g[pos][1] += g[son][1];
    }
  }
}
void dp(int pos, int dad, int dep) {
  int opt = dep & 1;
  for (unsigned i = 0, si = a[pos].size(); i < si; ++i) {
    int son = a[pos][i];
    if (son != dad) {
      dp(son, pos, dep + 1);
    }
  }
  ans = inc(ans, mul(w[pos], n));
  ans = inc(ans, mul(mul(w[pos], dec(g[1][opt], g[pos][opt])), size[pos]));
  ans = dec(ans,
            mul(mul(w[pos], dec(g[1][opt ^ 1], g[pos][opt ^ 1])), size[pos]));
  for (unsigned i = 0, si = a[pos].size(); i < si; ++i) {
    int son = a[pos][i];
    if (son != dad) {
      ans = inc(ans, mul(mul(w[pos], g[son][opt]), dec(size[1], size[son])));
      ans =
          dec(ans, mul(mul(w[pos], g[son][opt ^ 1]), dec(size[1], size[son])));
    }
  }
}
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) {
    read(w[i]);
    w[i] = w[i] % 1000000007;
    w[i] = (w[i] + 1000000007) % 1000000007;
  }
  for (int i = 1; i < n; ++i) {
    int u, v;
    read(u), read(v);
    a[u].push_back(v);
    a[v].push_back(u);
  }
  dfs(1, 0, 1);
  dp(1, 0, 1);
  printf("%d\n", ans);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int n, w[200010], size[200010], ans, g[200010][2];
vector<int> a[200010];
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
int inc(int x, int y) {
  return (x + y >= 1000000007) ? (x + y - 1000000007) : (x + y);
}
int dec(int x, int y) { return (x - y < 0) ? (x - y + 1000000007) : (x - y); }
int mul(int x, int y) { return 1ll * x * y % 1000000007; }
void dfs(int pos, int dad, int dep) {
  size[pos] = 1, g[pos][dep & 1] = 1;
  for (unsigned i = 0, si = a[pos].size(); i < si; ++i) {
    int son = a[pos][i];
    if (son != dad) {
      dfs(son, pos, dep + 1);
      size[pos] += size[son];
      g[pos][0] += g[son][0], g[pos][1] += g[son][1];
    }
  }
}
void dp(int pos, int dad, int dep) {
  int opt = dep & 1;
  for (unsigned i = 0, si = a[pos].size(); i < si; ++i) {
    int son = a[pos][i];
    if (son != dad) {
      dp(son, pos, dep + 1);
    }
  }
  ans = inc(ans, mul(w[pos], n));
  ans = inc(ans, mul(mul(w[pos], dec(g[1][opt], g[pos][opt])), size[pos]));
  ans = dec(ans,
            mul(mul(w[pos], dec(g[1][opt ^ 1], g[pos][opt ^ 1])), size[pos]));
  for (unsigned i = 0, si = a[pos].size(); i < si; ++i) {
    int son = a[pos][i];
    if (son != dad) {
      ans = inc(ans, mul(mul(w[pos], g[son][opt]), dec(size[1], size[son])));
      ans =
          dec(ans, mul(mul(w[pos], g[son][opt ^ 1]), dec(size[1], size[son])));
    }
  }
}
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) {
    read(w[i]);
    w[i] = w[i] % 1000000007;
    w[i] = (w[i] + 1000000007) % 1000000007;
  }
  for (int i = 1; i < n; ++i) {
    int u, v;
    read(u), read(v);
    a[u].push_back(v);
    a[v].push_back(u);
  }
  dfs(1, 0, 1);
  dp(1, 0, 1);
  printf("%d\n", ans);
  return 0;
}

```