#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7, P = 1e9 + 7;
struct edge {
  long long t, nxt;
} e[N << 1];
long long n, cnt, ans;
long long v[N], head[N], siz[N][2], f[N][2];
void add(long long u, long long t) {
  e[++cnt].t = t;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
void dfs(long long u, long long fa) {
  for (long long i = head[u]; i; i = e[i].nxt) {
    long long t = e[i].t;
    if (t == fa) continue;
    dfs(t, u);
    siz[u][1] += siz[t][0];
    siz[u][0] += siz[t][1];
    f[u][1] = (f[u][1] + f[t][0] + 1ll * v[u] * siz[t][0] % P + P) % P;
    f[u][0] = (f[u][0] + f[t][1] - 1ll * v[u] * siz[t][1] % P + P) % P;
  }
  for (int i = head[u]; i; i = e[i].nxt) {
    int t = e[i].t;
    if (t == fa) continue;
    if (siz[t][0])
      ans = (ans + 2ll * f[t][0] * (siz[u][1] - siz[t][0]) % P +
             siz[t][0] * (siz[u][1] - siz[t][0]) % P * v[u] % P) %
            P;
    if (siz[t][1])
      ans = (ans + 2ll * f[t][1] * (siz[u][0] - siz[t][1]) % P -
             siz[t][1] * (siz[u][0] - siz[t][1]) % P * v[u] % P) %
            P;
  }
  ans = (ans + f[u][1] * 2ll % P + v[u] + P) % P;
  siz[u][1]++;
  f[u][1] = (f[u][1] + v[u] + P) % P;
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &v[i]);
  for (long long i = 1, u, v; i < n; i++)
    scanf("%lld%lld", &u, &v), add(u, v), add(v, u);
  dfs(1, 0);
  printf("%lld\n", ans % P);
}

### 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;
const int N = 2e5 + 7, P = 1e9 + 7;
struct edge {
  long long t, nxt;
} e[N << 1];
long long n, cnt, ans;
long long v[N], head[N], siz[N][2], f[N][2];
void add(long long u, long long t) {
  e[++cnt].t = t;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
void dfs(long long u, long long fa) {
  for (long long i = head[u]; i; i = e[i].nxt) {
    long long t = e[i].t;
    if (t == fa) continue;
    dfs(t, u);
    siz[u][1] += siz[t][0];
    siz[u][0] += siz[t][1];
    f[u][1] = (f[u][1] + f[t][0] + 1ll * v[u] * siz[t][0] % P + P) % P;
    f[u][0] = (f[u][0] + f[t][1] - 1ll * v[u] * siz[t][1] % P + P) % P;
  }
  for (int i = head[u]; i; i = e[i].nxt) {
    int t = e[i].t;
    if (t == fa) continue;
    if (siz[t][0])
      ans = (ans + 2ll * f[t][0] * (siz[u][1] - siz[t][0]) % P +
             siz[t][0] * (siz[u][1] - siz[t][0]) % P * v[u] % P) %
            P;
    if (siz[t][1])
      ans = (ans + 2ll * f[t][1] * (siz[u][0] - siz[t][1]) % P -
             siz[t][1] * (siz[u][0] - siz[t][1]) % P * v[u] % P) %
            P;
  }
  ans = (ans + f[u][1] * 2ll % P + v[u] + P) % P;
  siz[u][1]++;
  f[u][1] = (f[u][1] + v[u] + P) % P;
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &v[i]);
  for (long long i = 1, u, v; i < n; i++)
    scanf("%lld%lld", &u, &v), add(u, v), add(v, u);
  dfs(1, 0);
  printf("%lld\n", ans % P);
}

```