#include <bits/stdc++.h>
using namespace std;
inline int Gcd(int X, int Y) { return Y ? Gcd(Y, X % Y) : X; }
inline long long Gcd(long long X, long long Y) { return Y ? Gcd(Y, X % Y) : X; }
inline int Pow(int base, long long exp, int _mod) {
  int _ans = 1;
  for (; exp; exp >>= 1, base = (long long)base * base % _mod)
    exp& 1 ? _ans = (long long)_ans * base % _mod : 0;
  return _ans;
}
inline long long Pow(long long base, long long exp, long long _mod) {
  long long _ans = 1;
  for (; exp; exp >>= 1, base = base * base % _mod)
    exp& 1 ? _ans = _ans * base % _mod : 0;
  return _ans;
}
const int INF = 0x3f3f3f3f;
int Mod = 1000000007;
int n, q;
int val[200001];
int h[200001], nxt[400001], to[400001], tot;
inline void ins(int x, int y) {
  nxt[++tot] = h[x];
  to[tot] = y;
  h[x] = tot;
}
int f[200001], siz[200001], dep[200001];
long long Ans;
void DFS(int u, int fa) {
  f[u] = 1;
  dep[u] = dep[fa] + 1;
  siz[u] = 1;
  for (int i = h[u]; i; i = nxt[i])
    if (to[i] != fa) DFS(to[i], u), f[u] -= f[to[i]], siz[u] += siz[to[i]];
}
void D2(int u, int fa) {
  for (int i = h[u]; i; i = nxt[i])
    if (to[i] != fa) {
      D2(to[i], u);
      Ans =
          ((Ans + (long long)f[to[i]] * siz[to[i]] % Mod * val[u] % Mod) % Mod +
           Mod) %
          Mod;
    }
  if (fa != 0)
    Ans = ((Ans + (long long)(((dep[u] & 1) ? -f[1] : f[1]) + f[u]) *
                      (n - siz[u]) % Mod * val[u] % Mod) %
               Mod +
           Mod) %
          Mod;
}
int main() {
  int x, y;
  scanf("%d", &n);
  for (int i = 1; i <= (n); ++i) scanf("%d", val + i);
  for (int i = 2; i <= (n); ++i) scanf("%d%d", &x, &y), ins(x, y), ins(y, x);
  DFS(1, 0);
  for (int i = 1; i <= (n); ++i)
    if (dep[i] & 1)
      Ans += (long long)f[1] * val[i];
    else
      Ans -= (long long)f[1] * val[i];
  Ans = (Ans % Mod + Mod) % Mod;
  Ans = Ans * n % Mod;
  D2(1, 0);
  printf("%lld\n", Ans);
  return 0;
}

### Prompt

Your task is to create 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;
inline int Gcd(int X, int Y) { return Y ? Gcd(Y, X % Y) : X; }
inline long long Gcd(long long X, long long Y) { return Y ? Gcd(Y, X % Y) : X; }
inline int Pow(int base, long long exp, int _mod) {
  int _ans = 1;
  for (; exp; exp >>= 1, base = (long long)base * base % _mod)
    exp& 1 ? _ans = (long long)_ans * base % _mod : 0;
  return _ans;
}
inline long long Pow(long long base, long long exp, long long _mod) {
  long long _ans = 1;
  for (; exp; exp >>= 1, base = base * base % _mod)
    exp& 1 ? _ans = _ans * base % _mod : 0;
  return _ans;
}
const int INF = 0x3f3f3f3f;
int Mod = 1000000007;
int n, q;
int val[200001];
int h[200001], nxt[400001], to[400001], tot;
inline void ins(int x, int y) {
  nxt[++tot] = h[x];
  to[tot] = y;
  h[x] = tot;
}
int f[200001], siz[200001], dep[200001];
long long Ans;
void DFS(int u, int fa) {
  f[u] = 1;
  dep[u] = dep[fa] + 1;
  siz[u] = 1;
  for (int i = h[u]; i; i = nxt[i])
    if (to[i] != fa) DFS(to[i], u), f[u] -= f[to[i]], siz[u] += siz[to[i]];
}
void D2(int u, int fa) {
  for (int i = h[u]; i; i = nxt[i])
    if (to[i] != fa) {
      D2(to[i], u);
      Ans =
          ((Ans + (long long)f[to[i]] * siz[to[i]] % Mod * val[u] % Mod) % Mod +
           Mod) %
          Mod;
    }
  if (fa != 0)
    Ans = ((Ans + (long long)(((dep[u] & 1) ? -f[1] : f[1]) + f[u]) *
                      (n - siz[u]) % Mod * val[u] % Mod) %
               Mod +
           Mod) %
          Mod;
}
int main() {
  int x, y;
  scanf("%d", &n);
  for (int i = 1; i <= (n); ++i) scanf("%d", val + i);
  for (int i = 2; i <= (n); ++i) scanf("%d%d", &x, &y), ins(x, y), ins(y, x);
  DFS(1, 0);
  for (int i = 1; i <= (n); ++i)
    if (dep[i] & 1)
      Ans += (long long)f[1] * val[i];
    else
      Ans -= (long long)f[1] * val[i];
  Ans = (Ans % Mod + Mod) % Mod;
  Ans = Ans * n % Mod;
  D2(1, 0);
  printf("%lld\n", Ans);
  return 0;
}

```