#include <bits/stdc++.h>
using namespace std;
void read(long long &first) {
  first = 0;
  long long f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    first = (first << 1) + (first << 3) + c - '0';
    c = getchar();
  }
  first *= f;
  return;
}
long long fs(long long first) { return first < 0 ? -first : first; }
void print(long long first) {
  if (first < 10) {
    putchar(first + '0');
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
  return;
}
void write(long long first) {
  if (first < 0) {
    putchar('-');
    first = fs(first);
  }
  print(first);
  return;
}
long long mint(long long first, long long second) {
  return first < second ? first : second;
}
long long maxt(long long first, long long second) {
  return first > second ? first : second;
}
vector<long long> G[200005];
long long ans, n, f[200005], g[200005], uf[200005], ug[200005], a[200005],
    siz[200005], fat[200005];
void dfs(long long u, long long fa) {
  f[u] = siz[u] = 1;
  fat[u] = fa;
  for (int i = 0; i < G[u].size(); i++) {
    long long v = G[u][i];
    if (v == fa) continue;
    dfs(v, u);
    f[u] += g[v];
    g[u] += f[v];
    siz[u] += siz[v];
  }
  ans = (ans + a[u] * (n - siz[u] + 1ll + 1000000007) % 1000000007 *
                   (f[u] - g[u] + 1000000007) % 1000000007) %
        1000000007;
  ans = (ans + a[u] * (siz[u] - 1ll + 1000000007) % 1000000007) % 1000000007;
  for (int i = 0; i < G[u].size(); i++) {
    long long v = G[u][i];
    if (v == fa) continue;
    ans = (ans + a[u] * (g[v] - f[v] + 1000000007) % 1000000007 *
                     (siz[u] - siz[v] - 1ll + 1000000007) % 1000000007) %
          1000000007;
  }
}
void bfs() {
  queue<long long> q;
  for (int i = 0; i < G[1].size(); i++) q.push(G[1][i]);
  while (!q.empty()) {
    long long p = q.front();
    q.pop();
    uf[p] = ug[fat[p]] + g[fat[p]] - f[p];
    ug[p] = uf[fat[p]] + f[fat[p]] - g[p];
    ans = (ans + ((uf[p] - ug[p] + 1000000007) % 1000000007 * siz[p] %
                  1000000007 * a[p] % 1000000007) %
                     1000000007) %
          1000000007;
    for (int i = 0; i < G[p].size(); i++)
      if (G[p][i] != fat[p]) q.push(G[p][i]);
  }
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i < n; i++) {
    long long u, v;
    read(u), read(v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 1);
  bfs();
  printf("%lld", (ans + 1000000007) % 1000000007);
}

### Prompt

Generate 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;
void read(long long &first) {
  first = 0;
  long long f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    first = (first << 1) + (first << 3) + c - '0';
    c = getchar();
  }
  first *= f;
  return;
}
long long fs(long long first) { return first < 0 ? -first : first; }
void print(long long first) {
  if (first < 10) {
    putchar(first + '0');
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
  return;
}
void write(long long first) {
  if (first < 0) {
    putchar('-');
    first = fs(first);
  }
  print(first);
  return;
}
long long mint(long long first, long long second) {
  return first < second ? first : second;
}
long long maxt(long long first, long long second) {
  return first > second ? first : second;
}
vector<long long> G[200005];
long long ans, n, f[200005], g[200005], uf[200005], ug[200005], a[200005],
    siz[200005], fat[200005];
void dfs(long long u, long long fa) {
  f[u] = siz[u] = 1;
  fat[u] = fa;
  for (int i = 0; i < G[u].size(); i++) {
    long long v = G[u][i];
    if (v == fa) continue;
    dfs(v, u);
    f[u] += g[v];
    g[u] += f[v];
    siz[u] += siz[v];
  }
  ans = (ans + a[u] * (n - siz[u] + 1ll + 1000000007) % 1000000007 *
                   (f[u] - g[u] + 1000000007) % 1000000007) %
        1000000007;
  ans = (ans + a[u] * (siz[u] - 1ll + 1000000007) % 1000000007) % 1000000007;
  for (int i = 0; i < G[u].size(); i++) {
    long long v = G[u][i];
    if (v == fa) continue;
    ans = (ans + a[u] * (g[v] - f[v] + 1000000007) % 1000000007 *
                     (siz[u] - siz[v] - 1ll + 1000000007) % 1000000007) %
          1000000007;
  }
}
void bfs() {
  queue<long long> q;
  for (int i = 0; i < G[1].size(); i++) q.push(G[1][i]);
  while (!q.empty()) {
    long long p = q.front();
    q.pop();
    uf[p] = ug[fat[p]] + g[fat[p]] - f[p];
    ug[p] = uf[fat[p]] + f[fat[p]] - g[p];
    ans = (ans + ((uf[p] - ug[p] + 1000000007) % 1000000007 * siz[p] %
                  1000000007 * a[p] % 1000000007) %
                     1000000007) %
          1000000007;
    for (int i = 0; i < G[p].size(); i++)
      if (G[p][i] != fat[p]) q.push(G[p][i]);
  }
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i < n; i++) {
    long long u, v;
    read(u), read(v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 1);
  bfs();
  printf("%lld", (ans + 1000000007) % 1000000007);
}

```