#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T& x) {
  char c = getchar();
  bool f = false;
  for (x = 0; !isdigit(c); c = getchar()) {
    if (c == '-') {
      f = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (f) {
    x = -x;
  }
}
template <typename T, typename... U>
inline void read(T& x, U&... y) {
  read(x), read(y...);
}
const int N = 2e5 + 10, P = 1e9 + 7;
int n;
int V[N], son[N][2], fa[N], num[N], top[N][2];
vector<int> E[N];
void DFS(int u) {
  son[u][1] = num[u] = 1;
  for (auto v : E[u])
    if (v != fa[u]) {
      fa[v] = u;
      DFS(v);
      num[u] += num[v];
    }
  son[fa[u]][0] += son[u][1];
  son[fa[u]][1] += son[u][0];
}
int DP(int u) {
  top[u][0] = top[fa[u]][1] + son[fa[u]][1] - son[u][0];
  top[u][1] = top[fa[u]][0] + son[fa[u]][0] - son[u][1];
  int res = 0;
  for (auto v : E[u])
    if (v != fa[u]) {
      (res += DP(v)) %= P;
      int t1 =
          1ll * (son[v][0] - son[v][1] + P) % P * (n - num[v]) % P * V[u] % P;
      (res += t1) %= P;
      int t2 = 1ll * (top[v][1] - top[v][0] + P) % P * num[v] % P * V[v] % P;
      (res += t2) %= P;
    }
  int t1 = 1ll * (n - num[u] + 1) * V[u] % P;
  (res += t1) %= P;
  int t2 = 1ll * (num[u] - 1) * V[u] % P;
  (res += t2) %= P;
  return res;
}
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) read(V[i]), V[i] = (V[i] % P + P) % P;
  for (int i = 1; i < n; ++i) {
    int u, v;
    read(u, v);
    E[u].push_back(v);
    E[v].push_back(u);
  }
  DFS(1);
  printf("%d\n", DP(1));
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 <typename T>
inline void read(T& x) {
  char c = getchar();
  bool f = false;
  for (x = 0; !isdigit(c); c = getchar()) {
    if (c == '-') {
      f = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (f) {
    x = -x;
  }
}
template <typename T, typename... U>
inline void read(T& x, U&... y) {
  read(x), read(y...);
}
const int N = 2e5 + 10, P = 1e9 + 7;
int n;
int V[N], son[N][2], fa[N], num[N], top[N][2];
vector<int> E[N];
void DFS(int u) {
  son[u][1] = num[u] = 1;
  for (auto v : E[u])
    if (v != fa[u]) {
      fa[v] = u;
      DFS(v);
      num[u] += num[v];
    }
  son[fa[u]][0] += son[u][1];
  son[fa[u]][1] += son[u][0];
}
int DP(int u) {
  top[u][0] = top[fa[u]][1] + son[fa[u]][1] - son[u][0];
  top[u][1] = top[fa[u]][0] + son[fa[u]][0] - son[u][1];
  int res = 0;
  for (auto v : E[u])
    if (v != fa[u]) {
      (res += DP(v)) %= P;
      int t1 =
          1ll * (son[v][0] - son[v][1] + P) % P * (n - num[v]) % P * V[u] % P;
      (res += t1) %= P;
      int t2 = 1ll * (top[v][1] - top[v][0] + P) % P * num[v] % P * V[v] % P;
      (res += t2) %= P;
    }
  int t1 = 1ll * (n - num[u] + 1) * V[u] % P;
  (res += t1) %= P;
  int t2 = 1ll * (num[u] - 1) * V[u] % P;
  (res += t2) %= P;
  return res;
}
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) read(V[i]), V[i] = (V[i] % P + P) % P;
  for (int i = 1; i < n; ++i) {
    int u, v;
    read(u, v);
    E[u].push_back(v);
    E[v].push_back(u);
  }
  DFS(1);
  printf("%d\n", DP(1));
  return 0;
}

```