#include <bits/stdc++.h>
using namespace std;
int get() {
  int x = 0, f = 1;
  char s;
  while ((s = getchar()) < '0' || s > '9')
    if (s == '-') f = -1;
  while (s >= '0' && s <= '9')
    x = (x << 3) + (x << 1) + (s ^ 48), s = getchar();
  return x * f;
}
const int M = 2e5 + 5, mo = 1e9 + 7;
int n, a[M], head[M], cntedge, fa[M], siz[M], f[M], g[M], ans, uf[M], ug[M];
struct edge {
  int to, nxt;
  edge() {}
  edge(const int T, const int NX) { to = T, nxt = NX; }
} e[M << 1];
void add(const int u, const int v) {
  e[++cntedge] = edge(v, head[u]), head[u] = cntedge;
}
void dfs(const int x) {
  f[x] = siz[x] = 1;
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to ^ fa[x]) {
      fa[e[i].to] = x;
      dfs(e[i].to);
      siz[x] += siz[e[i].to];
      f[x] += g[e[i].to];
      g[x] += f[e[i].to];
    }
  ans = (ans + 1ll * (f[x] - g[x]) * (n - siz[x] + 1) % mo * a[x] % mo) % mo;
  ans = (ans + 1ll * (siz[x] - 1) * a[x] % mo) % mo;
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to ^ fa[x])
      ans = (ans + 1ll * (g[e[i].to] - f[e[i].to]) *
                       (siz[x] - siz[e[i].to] - 1) % mo * a[x] % mo) %
            mo;
}
void bfs() {
  queue<int> q;
  for (int i = head[1]; i; i = e[i].nxt) q.push(e[i].to);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    uf[x] = ug[fa[x]] + g[fa[x]] - f[x];
    ug[x] = uf[fa[x]] + f[fa[x]] - g[x];
    ans = (ans + 1ll * (uf[x] - ug[x]) * siz[x] % mo * a[x] % mo) % mo;
    for (int i = head[x]; i; i = e[i].nxt)
      if (e[i].to ^ fa[x]) q.push(e[i].to);
  }
}
int main() {
  n = get();
  for (int i = 1; i <= n; ++i) a[i] = get();
  for (int i = 1, u, v; i < n; ++i) {
    u = get(), v = get();
    add(u, v), add(v, u);
  }
  dfs(1);
  bfs();
  printf("%d", (ans + mo) % mo);
  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;
int get() {
  int x = 0, f = 1;
  char s;
  while ((s = getchar()) < '0' || s > '9')
    if (s == '-') f = -1;
  while (s >= '0' && s <= '9')
    x = (x << 3) + (x << 1) + (s ^ 48), s = getchar();
  return x * f;
}
const int M = 2e5 + 5, mo = 1e9 + 7;
int n, a[M], head[M], cntedge, fa[M], siz[M], f[M], g[M], ans, uf[M], ug[M];
struct edge {
  int to, nxt;
  edge() {}
  edge(const int T, const int NX) { to = T, nxt = NX; }
} e[M << 1];
void add(const int u, const int v) {
  e[++cntedge] = edge(v, head[u]), head[u] = cntedge;
}
void dfs(const int x) {
  f[x] = siz[x] = 1;
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to ^ fa[x]) {
      fa[e[i].to] = x;
      dfs(e[i].to);
      siz[x] += siz[e[i].to];
      f[x] += g[e[i].to];
      g[x] += f[e[i].to];
    }
  ans = (ans + 1ll * (f[x] - g[x]) * (n - siz[x] + 1) % mo * a[x] % mo) % mo;
  ans = (ans + 1ll * (siz[x] - 1) * a[x] % mo) % mo;
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to ^ fa[x])
      ans = (ans + 1ll * (g[e[i].to] - f[e[i].to]) *
                       (siz[x] - siz[e[i].to] - 1) % mo * a[x] % mo) %
            mo;
}
void bfs() {
  queue<int> q;
  for (int i = head[1]; i; i = e[i].nxt) q.push(e[i].to);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    uf[x] = ug[fa[x]] + g[fa[x]] - f[x];
    ug[x] = uf[fa[x]] + f[fa[x]] - g[x];
    ans = (ans + 1ll * (uf[x] - ug[x]) * siz[x] % mo * a[x] % mo) % mo;
    for (int i = head[x]; i; i = e[i].nxt)
      if (e[i].to ^ fa[x]) q.push(e[i].to);
  }
}
int main() {
  n = get();
  for (int i = 1; i <= n; ++i) a[i] = get();
  for (int i = 1, u, v; i < n; ++i) {
    u = get(), v = get();
    add(u, v), add(v, u);
  }
  dfs(1);
  bfs();
  printf("%d", (ans + mo) % mo);
  return 0;
}

```