#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <class T>
inline bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
inline long long Max(long long x, long long y) { return x > y ? x : y; }
inline long long Min(long long x, long long y) { return x < y ? x : y; }
int n;
long long v[200002];
vector<int> E[200002];
long long sz[200002][2], dp[200002][2];
long long ans;
void Add(long long &x, long long y) {
  x += y % 1000000007;
  if (x >= 1000000007) x %= 1000000007;
  if (x < 0) (x += 1000000007) %= 1000000007;
}
void dfs(int x, int f) {
  dp[x][1] = v[x];
  sz[x][1] = 1;
  Add(ans, v[x]);
  for (int i = (0), i_end_ = (E[x].size()); i < i_end_; i++) {
    int y = E[x][i];
    if (y == f) continue;
    dfs(y, x);
    Add(ans, sz[x][0] * dp[y][1] % 1000000007 * 2 % 1000000007);
    Add(ans, dp[x][0] * sz[y][1] % 1000000007 * 2 % 1000000007);
    Add(ans, sz[x][1] * dp[y][0] % 1000000007 * 2 % 1000000007);
    Add(ans, dp[x][1] * sz[y][0] % 1000000007 * 2 % 1000000007);
    sz[x][0] += sz[y][1];
    sz[x][1] += sz[y][0];
    Add(dp[x][0], dp[y][1] - v[x] * sz[y][1] % 1000000007 + 1000000007);
    Add(dp[x][1], dp[y][0] + v[x] * sz[y][0] % 1000000007 + 1000000007);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = (1), i_end_ = (n); i <= i_end_; i++) {
    scanf("%lld", &v[i]);
    (v[i] += 1000000007) %= 1000000007;
  }
  for (int i = (1), i_end_ = (n); i < i_end_; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    E[a].push_back(b);
    E[b].push_back(a);
  }
  dfs(1, 0);
  printf("%lld\n", ans);
  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;
template <class T>
inline bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <class T>
inline bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
inline long long Max(long long x, long long y) { return x > y ? x : y; }
inline long long Min(long long x, long long y) { return x < y ? x : y; }
int n;
long long v[200002];
vector<int> E[200002];
long long sz[200002][2], dp[200002][2];
long long ans;
void Add(long long &x, long long y) {
  x += y % 1000000007;
  if (x >= 1000000007) x %= 1000000007;
  if (x < 0) (x += 1000000007) %= 1000000007;
}
void dfs(int x, int f) {
  dp[x][1] = v[x];
  sz[x][1] = 1;
  Add(ans, v[x]);
  for (int i = (0), i_end_ = (E[x].size()); i < i_end_; i++) {
    int y = E[x][i];
    if (y == f) continue;
    dfs(y, x);
    Add(ans, sz[x][0] * dp[y][1] % 1000000007 * 2 % 1000000007);
    Add(ans, dp[x][0] * sz[y][1] % 1000000007 * 2 % 1000000007);
    Add(ans, sz[x][1] * dp[y][0] % 1000000007 * 2 % 1000000007);
    Add(ans, dp[x][1] * sz[y][0] % 1000000007 * 2 % 1000000007);
    sz[x][0] += sz[y][1];
    sz[x][1] += sz[y][0];
    Add(dp[x][0], dp[y][1] - v[x] * sz[y][1] % 1000000007 + 1000000007);
    Add(dp[x][1], dp[y][0] + v[x] * sz[y][0] % 1000000007 + 1000000007);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = (1), i_end_ = (n); i <= i_end_; i++) {
    scanf("%lld", &v[i]);
    (v[i] += 1000000007) %= 1000000007;
  }
  for (int i = (1), i_end_ = (n); i < i_end_; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    E[a].push_back(b);
    E[b].push_back(a);
  }
  dfs(1, 0);
  printf("%lld\n", ans);
  return 0;
}

```