#include <bits/stdc++.h>
using namespace std;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' or c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' and c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
struct EDGE {
  int to, Next;
} edge[400100];
struct DOTS {
  long long cnt, num[2], all, sum, dpth;
  vector<int> sons;
} dot[200100];
int n;
int h[200100], cnt;
long long ans;
bool vis[200100];
void add(int from, int to) {
  edge[++cnt].to = to;
  edge[cnt].Next = h[from];
  h[from] = cnt;
}
void dfs(int t) {
  vis[t] = 1;
  dot[t].num[0]++;
  for (int i = h[t]; i; i = edge[i].Next) {
    int v = edge[i].to;
    if (!vis[v]) {
      dot[t].cnt++;
      dot[t].sons.push_back(v);
      dot[v].dpth = dot[t].dpth + 1;
      dfs(v);
      ans += dot[t].num[0] * dot[v].num[1] * 2 * dot[t].sum;
      ans %= 1000000007;
      ans -= dot[t].num[1] * dot[v].num[0] * 2 * dot[t].sum;
      ans %= 1000000007;
      dot[t].num[0] += dot[v].num[1];
      dot[t].num[1] += dot[v].num[0];
    }
  }
  ans += dot[t].sum;
}
void dp(int t) {
  for (int i = 0; i < dot[t].cnt; ++i) {
    int v = dot[t].sons[i];
    ans += dot[v].num[0] * (dot[1].num[dot[v].dpth & 1] - dot[v].num[0]) * 2 *
           dot[v].sum;
    ans %= 1000000007;
    ans -= dot[v].num[1] * (dot[1].num[(dot[v].dpth ^ 1) & 1] - dot[v].num[1]) *
           2 * dot[v].sum;
    ans %= 1000000007;
    dp(v);
  }
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) dot[i].sum = read();
  for (int i = 1; i < n; ++i) {
    int u = read();
    int v = read();
    add(u, v);
    add(v, u);
  }
  dfs(1);
  dp(1);
  ans %= 1000000007;
  ans += 1000000007;
  ans %= 1000000007;
  printf("%lld", ans);
  return 0;
}

### Prompt

Please create a solution in CPP 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;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' or c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' and c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
struct EDGE {
  int to, Next;
} edge[400100];
struct DOTS {
  long long cnt, num[2], all, sum, dpth;
  vector<int> sons;
} dot[200100];
int n;
int h[200100], cnt;
long long ans;
bool vis[200100];
void add(int from, int to) {
  edge[++cnt].to = to;
  edge[cnt].Next = h[from];
  h[from] = cnt;
}
void dfs(int t) {
  vis[t] = 1;
  dot[t].num[0]++;
  for (int i = h[t]; i; i = edge[i].Next) {
    int v = edge[i].to;
    if (!vis[v]) {
      dot[t].cnt++;
      dot[t].sons.push_back(v);
      dot[v].dpth = dot[t].dpth + 1;
      dfs(v);
      ans += dot[t].num[0] * dot[v].num[1] * 2 * dot[t].sum;
      ans %= 1000000007;
      ans -= dot[t].num[1] * dot[v].num[0] * 2 * dot[t].sum;
      ans %= 1000000007;
      dot[t].num[0] += dot[v].num[1];
      dot[t].num[1] += dot[v].num[0];
    }
  }
  ans += dot[t].sum;
}
void dp(int t) {
  for (int i = 0; i < dot[t].cnt; ++i) {
    int v = dot[t].sons[i];
    ans += dot[v].num[0] * (dot[1].num[dot[v].dpth & 1] - dot[v].num[0]) * 2 *
           dot[v].sum;
    ans %= 1000000007;
    ans -= dot[v].num[1] * (dot[1].num[(dot[v].dpth ^ 1) & 1] - dot[v].num[1]) *
           2 * dot[v].sum;
    ans %= 1000000007;
    dp(v);
  }
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) dot[i].sum = read();
  for (int i = 1; i < n; ++i) {
    int u = read();
    int v = read();
    add(u, v);
    add(v, u);
  }
  dfs(1);
  dp(1);
  ans %= 1000000007;
  ans += 1000000007;
  ans %= 1000000007;
  printf("%lld", ans);
  return 0;
}

```