#include <bits/stdc++.h>
using namespace std;
const int N = 202020, P = 1e9 + 7;
int n;
int a[N], par[N], cnt[2][N], c[2][N];
long long ans;
vector<int> g[N];
void upd(long long &a, long long b) {
  if (b < 0) b += P;
  a += b;
  if (a >= P) a -= P;
}
void dfs(int u, int fa) {
  cnt[0][u] = 1;
  cnt[1][u] = 0;
  par[u] = fa;
  for (int i = (0); i < ((int)g[u].size()); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    dfs(v, u);
    cnt[0][u] += cnt[1][v];
    cnt[1][u] += cnt[0][v];
  }
}
void dfs2(int u) {
  if (par[u]) {
    c[0][u] = c[1][par[u]];
    c[1][u] = c[0][par[u]];
  } else {
    c[0][u] = cnt[0][u];
    c[1][u] = cnt[1][u];
  }
  for (int i = (0); i < ((int)g[u].size()); i++) {
    int v = g[u][i];
    if (v == par[u]) continue;
    dfs2(v);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i < (n + 1); i++) {
    scanf("%d", a + i);
    if (a[i] < 0) a[i] += P;
  }
  for (int i = (1); i < (n); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  dfs2(1);
  for (int i = (1); i < (n + 1); i++) {
    long long pre = ans;
    int c0 = c[0][i], c1 = c[1][i];
    upd(ans, 1ll * c0 * c0 % P * a[i] % P);
    upd(ans, -1ll * c1 * c1 % P * a[i] % P);
    if (par[i]) {
      c0 = c[0][par[i]] - cnt[1][i];
      c1 = c[1][par[i]] - cnt[0][i];
      upd(ans, 1ll * c0 * c0 % P * a[i] % P);
      upd(ans, -1ll * c1 * c1 % P * a[i] % P);
    }
    for (int j = (0); j < ((int)g[i].size()); j++) {
      int v = g[i][j];
      if (v != par[i]) {
        c0 = cnt[0][v];
        c1 = cnt[1][v];
        upd(ans, 1ll * c0 * c0 % P * a[i] % P);
        upd(ans, -1ll * c1 * c1 % P * a[i] % P);
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### 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 = 202020, P = 1e9 + 7;
int n;
int a[N], par[N], cnt[2][N], c[2][N];
long long ans;
vector<int> g[N];
void upd(long long &a, long long b) {
  if (b < 0) b += P;
  a += b;
  if (a >= P) a -= P;
}
void dfs(int u, int fa) {
  cnt[0][u] = 1;
  cnt[1][u] = 0;
  par[u] = fa;
  for (int i = (0); i < ((int)g[u].size()); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    dfs(v, u);
    cnt[0][u] += cnt[1][v];
    cnt[1][u] += cnt[0][v];
  }
}
void dfs2(int u) {
  if (par[u]) {
    c[0][u] = c[1][par[u]];
    c[1][u] = c[0][par[u]];
  } else {
    c[0][u] = cnt[0][u];
    c[1][u] = cnt[1][u];
  }
  for (int i = (0); i < ((int)g[u].size()); i++) {
    int v = g[u][i];
    if (v == par[u]) continue;
    dfs2(v);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i < (n + 1); i++) {
    scanf("%d", a + i);
    if (a[i] < 0) a[i] += P;
  }
  for (int i = (1); i < (n); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  dfs2(1);
  for (int i = (1); i < (n + 1); i++) {
    long long pre = ans;
    int c0 = c[0][i], c1 = c[1][i];
    upd(ans, 1ll * c0 * c0 % P * a[i] % P);
    upd(ans, -1ll * c1 * c1 % P * a[i] % P);
    if (par[i]) {
      c0 = c[0][par[i]] - cnt[1][i];
      c1 = c[1][par[i]] - cnt[0][i];
      upd(ans, 1ll * c0 * c0 % P * a[i] % P);
      upd(ans, -1ll * c1 * c1 % P * a[i] % P);
    }
    for (int j = (0); j < ((int)g[i].size()); j++) {
      int v = g[i][j];
      if (v != par[i]) {
        c0 = cnt[0][v];
        c1 = cnt[1][v];
        upd(ans, 1ll * c0 * c0 % P * a[i] % P);
        upd(ans, -1ll * c1 * c1 % P * a[i] % P);
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```