#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
const int P = 1e9 + 7;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
int n, val[MAXN];
int size[MAXN], father[MAXN];
vector<int> a[MAXN];
int f[MAXN][2], g[MAXN][2];
void workg(int pos, int fa) {
  if (fa == 0)
    g[pos][0] = g[pos][1] = 0;
  else {
    g[pos][0] = f[fa][1] + g[fa][1] - f[pos][0];
    g[pos][1] = f[fa][0] + g[fa][0] - f[pos][1];
  }
  for (unsigned i = 0; i < a[pos].size(); i++)
    if (a[pos][i] != fa) workg(a[pos][i], pos);
}
void workf(int pos, int fa) {
  size[pos] = 1;
  father[pos] = fa;
  f[pos][0] = 1, f[pos][1] = 0;
  for (unsigned i = 0; i < a[pos].size(); i++)
    if (a[pos][i] != fa) {
      workf(a[pos][i], pos);
      size[pos] += size[a[pos][i]];
      f[pos][0] += f[a[pos][i]][1];
      f[pos][1] += f[a[pos][i]][0];
    }
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) {
    read(val[i]);
    if (val[i] < 0) val[i] += P;
  }
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    read(x), read(y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  workf(1, 0);
  workg(1, 0);
  int ans = 0;
  for (int pos = 1; pos <= n; pos++) {
    ans = (ans + 1ll * val[pos] * n) % P;
    for (unsigned i = 0; i < a[pos].size(); i++)
      if (a[pos][i] == father[pos]) {
        int lft = size[pos];
        ans = (ans + 1ll * lft * val[pos] % P * g[pos][0]) % P;
        ans = (ans + 1ll * lft * (P - val[pos]) % P * g[pos][1]) % P;
      } else {
        int lft = n - size[a[pos][i]];
        ans = (ans + 1ll * lft * val[pos] % P * f[a[pos][i]][1]) % P;
        ans = (ans + 1ll * lft * (P - val[pos]) % P * f[a[pos][i]][0]) % P;
      }
  }
  writeln(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;
const int MAXN = 200005;
const int P = 1e9 + 7;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
int n, val[MAXN];
int size[MAXN], father[MAXN];
vector<int> a[MAXN];
int f[MAXN][2], g[MAXN][2];
void workg(int pos, int fa) {
  if (fa == 0)
    g[pos][0] = g[pos][1] = 0;
  else {
    g[pos][0] = f[fa][1] + g[fa][1] - f[pos][0];
    g[pos][1] = f[fa][0] + g[fa][0] - f[pos][1];
  }
  for (unsigned i = 0; i < a[pos].size(); i++)
    if (a[pos][i] != fa) workg(a[pos][i], pos);
}
void workf(int pos, int fa) {
  size[pos] = 1;
  father[pos] = fa;
  f[pos][0] = 1, f[pos][1] = 0;
  for (unsigned i = 0; i < a[pos].size(); i++)
    if (a[pos][i] != fa) {
      workf(a[pos][i], pos);
      size[pos] += size[a[pos][i]];
      f[pos][0] += f[a[pos][i]][1];
      f[pos][1] += f[a[pos][i]][0];
    }
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) {
    read(val[i]);
    if (val[i] < 0) val[i] += P;
  }
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    read(x), read(y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  workf(1, 0);
  workg(1, 0);
  int ans = 0;
  for (int pos = 1; pos <= n; pos++) {
    ans = (ans + 1ll * val[pos] * n) % P;
    for (unsigned i = 0; i < a[pos].size(); i++)
      if (a[pos][i] == father[pos]) {
        int lft = size[pos];
        ans = (ans + 1ll * lft * val[pos] % P * g[pos][0]) % P;
        ans = (ans + 1ll * lft * (P - val[pos]) % P * g[pos][1]) % P;
      } else {
        int lft = n - size[a[pos][i]];
        ans = (ans + 1ll * lft * val[pos] % P * f[a[pos][i]][1]) % P;
        ans = (ans + 1ll * lft * (P - val[pos]) % P * f[a[pos][i]][0]) % P;
      }
  }
  writeln(ans);
  return 0;
}

```