#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, P = 1e9 + 7;
template <typename T>
inline void read(T &x) {
  char c;
  while (isspace(c = getchar()))
    ;
  bool neg = false;
  if (!isdigit(c))
    neg = (c == '-'), x = 0;
  else
    x = c - '0';
  while (isdigit(c = getchar())) x = x * 10 + c - '0';
  if (neg) x = -x;
}
struct par {
  par() : A(0), B(0) {}
  int A, B;
};
int fr[N], nx[N << 1], to[N << 1], E;
inline void adde(int u, int v) {
  to[++E] = v;
  nx[E] = fr[u];
  fr[u] = E;
}
long long total, sum;
int szA[N], szB[N], w[N];
int n;
inline void pls(int &a, const int &b) { a += b, a -= (a >= P) ? P : 0; }
inline void sub(int &a, const int &b) { a -= b, a += (a < 0) ? P : 0; }
par DFS(int x, int f) {
  par temp;
  szA[x] = 1;
  temp.A = w[x];
  sum += w[x];
  for (int i = fr[x], v; i; i = nx[i])
    if ((v = to[i]) != f) {
      par tmp = DFS(v, x);
      total += (1ll * tmp.A * szB[x] % P) + (1ll * temp.B * szA[v] % P) +
               (1ll * tmp.B * szA[x] % P) + (1ll * temp.A * szB[v] % P);
      total %= P;
      pls(temp.A, tmp.B);
      pls(temp.A, 1ll * szB[v] * w[x] % P);
      pls(temp.B, tmp.A);
      sub(temp.B, 1ll * szA[v] * w[x] % P);
      szA[x] += szB[v];
      szB[x] += szA[v];
    }
  return temp;
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(w[i]), pls(w[i], P);
  for (int i = 1, u, v; i < n; i++) {
    read(u);
    read(v);
    adde(u, v);
    adde(v, u);
  }
  DFS(1, 0);
  printf("%lld", ((total << 1) + sum) % P);
  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 N = 2e5 + 10, P = 1e9 + 7;
template <typename T>
inline void read(T &x) {
  char c;
  while (isspace(c = getchar()))
    ;
  bool neg = false;
  if (!isdigit(c))
    neg = (c == '-'), x = 0;
  else
    x = c - '0';
  while (isdigit(c = getchar())) x = x * 10 + c - '0';
  if (neg) x = -x;
}
struct par {
  par() : A(0), B(0) {}
  int A, B;
};
int fr[N], nx[N << 1], to[N << 1], E;
inline void adde(int u, int v) {
  to[++E] = v;
  nx[E] = fr[u];
  fr[u] = E;
}
long long total, sum;
int szA[N], szB[N], w[N];
int n;
inline void pls(int &a, const int &b) { a += b, a -= (a >= P) ? P : 0; }
inline void sub(int &a, const int &b) { a -= b, a += (a < 0) ? P : 0; }
par DFS(int x, int f) {
  par temp;
  szA[x] = 1;
  temp.A = w[x];
  sum += w[x];
  for (int i = fr[x], v; i; i = nx[i])
    if ((v = to[i]) != f) {
      par tmp = DFS(v, x);
      total += (1ll * tmp.A * szB[x] % P) + (1ll * temp.B * szA[v] % P) +
               (1ll * tmp.B * szA[x] % P) + (1ll * temp.A * szB[v] % P);
      total %= P;
      pls(temp.A, tmp.B);
      pls(temp.A, 1ll * szB[v] * w[x] % P);
      pls(temp.B, tmp.A);
      sub(temp.B, 1ll * szA[v] * w[x] % P);
      szA[x] += szB[v];
      szB[x] += szA[v];
    }
  return temp;
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(w[i]), pls(w[i], P);
  for (int i = 1, u, v; i < n; i++) {
    read(u);
    read(v);
    adde(u, v);
    adde(v, u);
  }
  DFS(1, 0);
  printf("%lld", ((total << 1) + sum) % P);
  return 0;
}

```