#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> adj[202020];
long long val[202020], qpar[202020], qimpar[202020], sobe[202020],
    desce[202020], ans, mod = 1e9 + 7;
void add(long long& a, long long b) { a = ((a + b) % mod + mod) % mod; }
long long mul(long long a, long long b) { return (a * b) % mod; }
void dfs(int v, int p) {
  long long totsobe = 0, totdesce = 0, sumpar = 0, sumimpar = 0;
  for (auto i = (0); i < (((int)(adj[v]).size())); i++)
    if (p != adj[v][i]) {
      int u = adj[v][i];
      dfs(u, v);
      add(qpar[v], qimpar[u]);
      add(qimpar[v], qpar[u]);
      add(ans, sobe[v] * (qpar[u] + qimpar[u]));
      add(ans, desce[u] * (sumpar - sumimpar));
      add(ans,
          (sobe[u] + mul(-qpar[u] + qimpar[u], val[v])) * (sumpar + sumimpar));
      add(ans, totdesce * (qpar[u] - qimpar[u]));
      add(sobe[v], sobe[u] + (-qpar[u] + qimpar[u]) * val[v]);
      add(desce[v], -desce[u] + (qpar[u] + qimpar[u]) * val[v]);
      add(totsobe, sobe[u]);
      add(totdesce, desce[u]);
      add(sumpar, qpar[u]);
      add(sumimpar, qimpar[u]);
    }
  add(qpar[v], 1);
  add(desce[v], val[v]);
  add(sobe[v], val[v]);
  add(ans, desce[v]);
  add(ans, sobe[v]);
  add(ans, -val[v]);
}
int main() {
  scanf("%d", &n);
  for (auto i = (0); i < (n); i++) scanf("%lld", val + i), add(val[i], mod);
  for (auto i = (1); i < (n); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--, v--;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  dfs(0, 0);
  printf("%lld\n", ans);
}

### Prompt

Your challenge is to write 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;
int n;
vector<int> adj[202020];
long long val[202020], qpar[202020], qimpar[202020], sobe[202020],
    desce[202020], ans, mod = 1e9 + 7;
void add(long long& a, long long b) { a = ((a + b) % mod + mod) % mod; }
long long mul(long long a, long long b) { return (a * b) % mod; }
void dfs(int v, int p) {
  long long totsobe = 0, totdesce = 0, sumpar = 0, sumimpar = 0;
  for (auto i = (0); i < (((int)(adj[v]).size())); i++)
    if (p != adj[v][i]) {
      int u = adj[v][i];
      dfs(u, v);
      add(qpar[v], qimpar[u]);
      add(qimpar[v], qpar[u]);
      add(ans, sobe[v] * (qpar[u] + qimpar[u]));
      add(ans, desce[u] * (sumpar - sumimpar));
      add(ans,
          (sobe[u] + mul(-qpar[u] + qimpar[u], val[v])) * (sumpar + sumimpar));
      add(ans, totdesce * (qpar[u] - qimpar[u]));
      add(sobe[v], sobe[u] + (-qpar[u] + qimpar[u]) * val[v]);
      add(desce[v], -desce[u] + (qpar[u] + qimpar[u]) * val[v]);
      add(totsobe, sobe[u]);
      add(totdesce, desce[u]);
      add(sumpar, qpar[u]);
      add(sumimpar, qimpar[u]);
    }
  add(qpar[v], 1);
  add(desce[v], val[v]);
  add(sobe[v], val[v]);
  add(ans, desce[v]);
  add(ans, sobe[v]);
  add(ans, -val[v]);
}
int main() {
  scanf("%d", &n);
  for (auto i = (0); i < (n); i++) scanf("%lld", val + i), add(val[i], mod);
  for (auto i = (1); i < (n); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--, v--;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  dfs(0, 0);
  printf("%lld\n", ans);
}

```