#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
const int N = 2e5 + 10, p = 1e9 + 7;
int las[N], to[N << 1], nxt[N << 1], cnt;
inline void add(int u, int v) {
  cnt++;
  to[cnt] = v;
  nxt[cnt] = las[u];
  las[u] = cnt;
}
inline void add2() {
  int a, b;
  scanf("%d%d", &a, &b);
  add(a, b);
  add(b, a);
}
int n, ans, rt, fa[N], val[N], siz[N], tot[N][2];
void dfs(int now, int la, int x) {
  int i;
  fa[now] = la;
  siz[now] = tot[now][x] = 1;
  for (i = las[now]; i; i = nxt[i])
    if (to[i] != la)
      dfs(to[i], now, x ^ 1), siz[now] += siz[to[i]],
          tot[now][x] += tot[to[i]][x], tot[now][x ^ 1] += tot[to[i]][x ^ 1];
}
void dfs(int now, int x) {
  int i;
  for (i = las[now]; i; i = nxt[i])
    if (to[i] != fa[now])
      dfs(to[i], x ^ 1),
          ans = (ans + (long long)(tot[to[i]][x] - tot[to[i]][x ^ 1] + p) % p *
                           (n - siz[to[i]]) % p * val[now] % p) %
                p;
  ans = (ans + (long long)((tot[rt][x] - tot[now][x]) -
                           (tot[rt][x ^ 1] - tot[now][x ^ 1]) + p) %
                   p * siz[now] % p * val[now] % p) %
        p;
  ans = (ans + (long long)n * val[now] % p) % p;
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &val[i]), val[i] = (val[i] + p) % p;
  for (i = 1; i < n; i++) add2();
  dfs(rt = 1, 0, 0);
  dfs(rt, 0);
  printf("%d\n", ans);
  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>
#pragma GCC optimize("Ofast")
const int N = 2e5 + 10, p = 1e9 + 7;
int las[N], to[N << 1], nxt[N << 1], cnt;
inline void add(int u, int v) {
  cnt++;
  to[cnt] = v;
  nxt[cnt] = las[u];
  las[u] = cnt;
}
inline void add2() {
  int a, b;
  scanf("%d%d", &a, &b);
  add(a, b);
  add(b, a);
}
int n, ans, rt, fa[N], val[N], siz[N], tot[N][2];
void dfs(int now, int la, int x) {
  int i;
  fa[now] = la;
  siz[now] = tot[now][x] = 1;
  for (i = las[now]; i; i = nxt[i])
    if (to[i] != la)
      dfs(to[i], now, x ^ 1), siz[now] += siz[to[i]],
          tot[now][x] += tot[to[i]][x], tot[now][x ^ 1] += tot[to[i]][x ^ 1];
}
void dfs(int now, int x) {
  int i;
  for (i = las[now]; i; i = nxt[i])
    if (to[i] != fa[now])
      dfs(to[i], x ^ 1),
          ans = (ans + (long long)(tot[to[i]][x] - tot[to[i]][x ^ 1] + p) % p *
                           (n - siz[to[i]]) % p * val[now] % p) %
                p;
  ans = (ans + (long long)((tot[rt][x] - tot[now][x]) -
                           (tot[rt][x ^ 1] - tot[now][x ^ 1]) + p) %
                   p * siz[now] % p * val[now] % p) %
        p;
  ans = (ans + (long long)n * val[now] % p) % p;
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &val[i]), val[i] = (val[i] + p) % p;
  for (i = 1; i < n; i++) add2();
  dfs(rt = 1, 0, 0);
  dfs(rt, 0);
  printf("%d\n", ans);
  return 0;
}

```