#include <bits/stdc++.h>
const int N = 2e5 + 5, mod = 1e9 + 7;
int n, a[N], nxt[N << 1], head[N], to[N << 1], cnt, ans, f[N], g[N], uf[N],
    ug[N], siz[N];
int read() {
  int x = 0, f = 1;
  char s;
  while ((s = getchar()) > '9' || s < '0')
    if (s == '-') f = -1;
  while (s >= '0' && s <= '9')
    x = (x << 1) + (x << 3) + (s ^ 48), s = getchar();
  return x * f;
}
void addEdge(const int u, const int v) {
  nxt[++cnt] = head[u], to[cnt] = v, head[u] = cnt;
}
int inc(int x, const int y) {
  x += y;
  if (x >= mod) return x - mod;
  if (x < 0) return x + mod;
  return x;
}
void DP1(const int u, const int fa) {
  siz[u] = f[u] = 1;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == fa) continue;
    DP1(v, u);
    f[u] += g[v], g[u] += f[v], siz[u] += siz[v];
  }
  ans = inc(ans, 1ll * a[u] * (f[u] - g[u]) % mod * (n - siz[u] + 1) % mod);
  ans = inc(ans, 1ll * a[u] * (siz[u] - 1) % mod);
  for (int i = head[u]; i; i = nxt[i])
    if (to[i] != fa)
      ans = inc(ans, 1ll * a[u] * (g[to[i]] - f[to[i]]) % mod *
                         (siz[u] - siz[to[i]] - 1) % mod);
}
void DP2(const int u, const int fa) {
  uf[u] = ug[fa] + g[fa] - f[u];
  ug[u] = uf[fa] + f[fa] - g[u];
  ans = inc(ans, 1ll * a[u] * (uf[u] - ug[u]) % mod * siz[u] % mod);
  for (int i = head[u]; i; i = nxt[i])
    if (to[i] != fa) DP2(to[i], u);
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  int u, v;
  for (int i = 1; i < n; ++i) {
    u = read(), v = read();
    addEdge(u, v), addEdge(v, u);
  }
  DP1(1, 0);
  uf[0] = g[1], ug[0] = f[1];
  DP2(1, 0);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
const int N = 2e5 + 5, mod = 1e9 + 7;
int n, a[N], nxt[N << 1], head[N], to[N << 1], cnt, ans, f[N], g[N], uf[N],
    ug[N], siz[N];
int read() {
  int x = 0, f = 1;
  char s;
  while ((s = getchar()) > '9' || s < '0')
    if (s == '-') f = -1;
  while (s >= '0' && s <= '9')
    x = (x << 1) + (x << 3) + (s ^ 48), s = getchar();
  return x * f;
}
void addEdge(const int u, const int v) {
  nxt[++cnt] = head[u], to[cnt] = v, head[u] = cnt;
}
int inc(int x, const int y) {
  x += y;
  if (x >= mod) return x - mod;
  if (x < 0) return x + mod;
  return x;
}
void DP1(const int u, const int fa) {
  siz[u] = f[u] = 1;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == fa) continue;
    DP1(v, u);
    f[u] += g[v], g[u] += f[v], siz[u] += siz[v];
  }
  ans = inc(ans, 1ll * a[u] * (f[u] - g[u]) % mod * (n - siz[u] + 1) % mod);
  ans = inc(ans, 1ll * a[u] * (siz[u] - 1) % mod);
  for (int i = head[u]; i; i = nxt[i])
    if (to[i] != fa)
      ans = inc(ans, 1ll * a[u] * (g[to[i]] - f[to[i]]) % mod *
                         (siz[u] - siz[to[i]] - 1) % mod);
}
void DP2(const int u, const int fa) {
  uf[u] = ug[fa] + g[fa] - f[u];
  ug[u] = uf[fa] + f[fa] - g[u];
  ans = inc(ans, 1ll * a[u] * (uf[u] - ug[u]) % mod * siz[u] % mod);
  for (int i = head[u]; i; i = nxt[i])
    if (to[i] != fa) DP2(to[i], u);
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  int u, v;
  for (int i = 1; i < n; ++i) {
    u = read(), v = read();
    addEdge(u, v), addEdge(v, u);
  }
  DP1(1, 0);
  uf[0] = g[1], ug[0] = f[1];
  DP2(1, 0);
  printf("%d\n", ans);
  return 0;
}

```