#include <bits/stdc++.h>
using namespace std;
const int mxn = 2e5 + 100, mod = 1e9 + 7;
int eu[2 * mxn], st[mxn], ne[2 * mxn], a[mxn];
long long ans = 0;
long long check(long long x) {
  if (x < 0) x += mod;
  if (x >= mod) x %= mod;
  return x;
}
long long add(long long x, long long y) { return check(check(x) + check(y)); }
long long mult(long long x, long long y) {
  return check(1LL * check(x) * check(y));
}
void merge_add(int au, vector<long long>& x, vector<long long>& y) {
  for (int i : {2, 6}) x[i] = check(-x[i]), y[i] = check(-y[i]);
  ans = add(
      ans, add(mult(x[0] + x[2], y[5] + y[7]), mult(x[1] + x[3], y[4] + y[6])));
  ans = add(
      ans, add(mult(x[5] + x[7], y[0] + y[2]), mult(x[4] + x[6], y[1] + y[3])));
  for (int i : {2, 6}) x[i] = check(-x[i]), y[i] = check(-y[i]);
  x[0] = add(x[0], add(y[4], mult(au, y[5]))), x[1] = add(x[1], y[5]);
  x[2] = add(x[2], add(y[6], mult(au, y[7]))), x[3] = add(x[3], y[7]);
  x[4] = add(x[4], add(y[0], mult(au, y[1]))), x[5] = add(x[5], y[1]);
  x[6] = add(x[6], add(y[2], mult(au, y[3]))), x[7] = add(x[7], y[3]);
}
vector<long long> dfs(int u, int p, int h) {
  vector<long long> in(8, 0);
  int s = h & 1 ? 1 : -1;
  for (int i = st[u]; i; i = ne[i])
    if (eu[i] != p) {
      vector<long long> it = dfs(eu[i], u, h + 1);
      merge_add(check(a[u] * s), in, it);
    }
  if (s == 1)
    in[4] = add(in[4], a[u]), in[5] = add(in[5], 1);
  else
    in[6] = add(in[6], -a[u]), in[7] = add(in[7], 1);
  ans = add(ans, add(in[4], -in[6]));
  return in;
}
int main() {
  int u, v, n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), a[i] = check(a[i]);
  for (int i = 1, s = 0; i < n; ++i) {
    scanf("%d%d", &u, &v);
    eu[++s] = v, ne[s] = st[u], st[u] = s;
    eu[++s] = u, ne[s] = st[v], st[v] = s;
  }
  dfs(1, 1, 1);
  ans = add(ans, ans);
  for (int i = 1; i <= n; i++) ans = add(ans, -a[i]);
  printf("%lld\n", ans);
}

### Prompt

Create a solution in Cpp for 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 int mxn = 2e5 + 100, mod = 1e9 + 7;
int eu[2 * mxn], st[mxn], ne[2 * mxn], a[mxn];
long long ans = 0;
long long check(long long x) {
  if (x < 0) x += mod;
  if (x >= mod) x %= mod;
  return x;
}
long long add(long long x, long long y) { return check(check(x) + check(y)); }
long long mult(long long x, long long y) {
  return check(1LL * check(x) * check(y));
}
void merge_add(int au, vector<long long>& x, vector<long long>& y) {
  for (int i : {2, 6}) x[i] = check(-x[i]), y[i] = check(-y[i]);
  ans = add(
      ans, add(mult(x[0] + x[2], y[5] + y[7]), mult(x[1] + x[3], y[4] + y[6])));
  ans = add(
      ans, add(mult(x[5] + x[7], y[0] + y[2]), mult(x[4] + x[6], y[1] + y[3])));
  for (int i : {2, 6}) x[i] = check(-x[i]), y[i] = check(-y[i]);
  x[0] = add(x[0], add(y[4], mult(au, y[5]))), x[1] = add(x[1], y[5]);
  x[2] = add(x[2], add(y[6], mult(au, y[7]))), x[3] = add(x[3], y[7]);
  x[4] = add(x[4], add(y[0], mult(au, y[1]))), x[5] = add(x[5], y[1]);
  x[6] = add(x[6], add(y[2], mult(au, y[3]))), x[7] = add(x[7], y[3]);
}
vector<long long> dfs(int u, int p, int h) {
  vector<long long> in(8, 0);
  int s = h & 1 ? 1 : -1;
  for (int i = st[u]; i; i = ne[i])
    if (eu[i] != p) {
      vector<long long> it = dfs(eu[i], u, h + 1);
      merge_add(check(a[u] * s), in, it);
    }
  if (s == 1)
    in[4] = add(in[4], a[u]), in[5] = add(in[5], 1);
  else
    in[6] = add(in[6], -a[u]), in[7] = add(in[7], 1);
  ans = add(ans, add(in[4], -in[6]));
  return in;
}
int main() {
  int u, v, n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), a[i] = check(a[i]);
  for (int i = 1, s = 0; i < n; ++i) {
    scanf("%d%d", &u, &v);
    eu[++s] = v, ne[s] = st[u], st[u] = s;
    eu[++s] = u, ne[s] = st[v], st[v] = s;
  }
  dfs(1, 1, 1);
  ans = add(ans, ans);
  for (int i = 1; i <= n; i++) ans = add(ans, -a[i]);
  printf("%lld\n", ans);
}

```