#include <bits/stdc++.h>
using namespace std;
inline void read(int &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
const int N = 500000, mo = 1e9 + 7;
int n, head[N], v[N], to[N], nxt[N], edgenum;
void add(int u, int v) {
  to[++edgenum] = v;
  nxt[edgenum] = head[u];
  head[u] = edgenum;
}
int size[N], tot, f[N], dep[N];
long long ans;
void dfs(int u, int fa, int depth) {
  size[u] = 1;
  dep[u] = depth;
  f[u] = (dep[u] == 1) ? 1 : -1;
  for (int i = head[u]; i != 0; i = nxt[i])
    if (to[i] != fa)
      dfs(to[i], u, depth ^ 1), size[u] += size[to[i]], f[u] += f[to[i]];
}
long long CNT[N];
void dfs1(int u, int fa) {
  int cnt = tot - (dep[u] == 1 ? 1 : -1);
  for (int i = head[u]; i != 0; i = nxt[i])
    if (to[i] != fa) {
      dfs1(to[i], u);
      int now = f[to[i]];
      cnt -= now;
      ans += 1LL * (dep[u] == 1 ? 1 : -1) * now * (n - size[to[i]]) % mo *
             v[u] % mo;
      CNT[u] += 1LL * (dep[u] == 1 ? 1 : -1) * now * (n - size[to[i]]) % mo;
    }
  CNT[u] += 1LL * (dep[u] == 1 ? 1 : -1) * cnt * (size[u]) % mo + n;
  ans += 1LL * (dep[u] == 1 ? 1 : -1) * cnt * (size[u]) % mo * v[u] % mo;
  ans += 1LL * n * v[u] % mo;
}
int main() {
  read(n);
  for (register int i = 1; i <= n; i++) read(v[i]);
  for (register int i = 2; i <= n; i++) {
    int x, y;
    read(x);
    read(y);
    add(x, y);
    add(y, x);
  }
  dfs(1, 0, 0);
  tot = f[1];
  dfs1(1, 0);
  cout << ((ans % mo + mo) % mo);
  return 0;
}

### 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;
inline void read(int &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
const int N = 500000, mo = 1e9 + 7;
int n, head[N], v[N], to[N], nxt[N], edgenum;
void add(int u, int v) {
  to[++edgenum] = v;
  nxt[edgenum] = head[u];
  head[u] = edgenum;
}
int size[N], tot, f[N], dep[N];
long long ans;
void dfs(int u, int fa, int depth) {
  size[u] = 1;
  dep[u] = depth;
  f[u] = (dep[u] == 1) ? 1 : -1;
  for (int i = head[u]; i != 0; i = nxt[i])
    if (to[i] != fa)
      dfs(to[i], u, depth ^ 1), size[u] += size[to[i]], f[u] += f[to[i]];
}
long long CNT[N];
void dfs1(int u, int fa) {
  int cnt = tot - (dep[u] == 1 ? 1 : -1);
  for (int i = head[u]; i != 0; i = nxt[i])
    if (to[i] != fa) {
      dfs1(to[i], u);
      int now = f[to[i]];
      cnt -= now;
      ans += 1LL * (dep[u] == 1 ? 1 : -1) * now * (n - size[to[i]]) % mo *
             v[u] % mo;
      CNT[u] += 1LL * (dep[u] == 1 ? 1 : -1) * now * (n - size[to[i]]) % mo;
    }
  CNT[u] += 1LL * (dep[u] == 1 ? 1 : -1) * cnt * (size[u]) % mo + n;
  ans += 1LL * (dep[u] == 1 ? 1 : -1) * cnt * (size[u]) % mo * v[u] % mo;
  ans += 1LL * n * v[u] % mo;
}
int main() {
  read(n);
  for (register int i = 1; i <= n; i++) read(v[i]);
  for (register int i = 2; i <= n; i++) {
    int x, y;
    read(x);
    read(y);
    add(x, y);
    add(y, x);
  }
  dfs(1, 0, 0);
  tot = f[1];
  dfs1(1, 0);
  cout << ((ans % mo + mo) % mo);
  return 0;
}

```