#include <bits/stdc++.h>
using namespace std;
long long int a[200052], num[200052][2];
vector<int> v[200052];
bool vis[200052];
int id[200052], fa[200052];
long long int n, s[200052];
void Get_id(int x) {
  vis[x] = 1;
  s[x] = 1;
  for (int i = 0; i < v[x].size(); i++) {
    if (vis[v[x][i]]) continue;
    fa[v[x][i]] = x;
    id[v[x][i]] = id[x] ^ 1;
    Get_id(v[x][i]);
    s[x] += s[v[x][i]];
  }
}
void dfs(int x) {
  long long int ret = 0;
  num[x][0] = num[x][1] = 0;
  num[x][id[x]]++;
  vis[x] = 1;
  for (int i = 0; i < v[x].size(); i++) {
    if (vis[v[x][i]]) continue;
    dfs(v[x][i]);
    num[x][0] += num[v[x][i]][0];
    num[x][1] += num[v[x][i]][1];
  }
}
long long int cla(int x) {
  long long int ret = 0;
  int t, k = id[x];
  for (int i = 0; i < v[x].size(); i++) {
    t = v[x][i];
    if (t == fa[x]) continue;
    ret = ((ret + ((((num[t][k] - num[t][k ^ 1]) % 1000000007) * (n - s[t])) %
                   1000000007)) %
           1000000007);
  }
  {
    ret =
        ((ret + (((((num[1][k] - num[x][k]) - (num[1][k ^ 1] - num[x][k ^ 1])) %
                   1000000007) *
                  s[x]) %
                 1000000007)) %
         1000000007);
    ret = ((ret + n) % 1000000007);
  }
  ret = ((((ret) % 1000000007) * ((a[x]) % 1000000007)) % 1000000007);
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  for (int i = 1; i < n; i++) {
    int ta, tb;
    scanf("%d%d", &ta, &tb);
    v[ta].push_back(tb);
    v[tb].push_back(ta);
  }
  memset(vis, 0, sizeof(vis));
  memset(num, 0, sizeof(num));
  id[1] = 1;
  fa[1] = 0;
  Get_id(1);
  memset(vis, 0, sizeof(vis));
  dfs(1);
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = ((ans + cla(i)) % 1000000007);
  }
  printf("%lld\n", ((ans + 1000000007) % 1000000007));
  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;
long long int a[200052], num[200052][2];
vector<int> v[200052];
bool vis[200052];
int id[200052], fa[200052];
long long int n, s[200052];
void Get_id(int x) {
  vis[x] = 1;
  s[x] = 1;
  for (int i = 0; i < v[x].size(); i++) {
    if (vis[v[x][i]]) continue;
    fa[v[x][i]] = x;
    id[v[x][i]] = id[x] ^ 1;
    Get_id(v[x][i]);
    s[x] += s[v[x][i]];
  }
}
void dfs(int x) {
  long long int ret = 0;
  num[x][0] = num[x][1] = 0;
  num[x][id[x]]++;
  vis[x] = 1;
  for (int i = 0; i < v[x].size(); i++) {
    if (vis[v[x][i]]) continue;
    dfs(v[x][i]);
    num[x][0] += num[v[x][i]][0];
    num[x][1] += num[v[x][i]][1];
  }
}
long long int cla(int x) {
  long long int ret = 0;
  int t, k = id[x];
  for (int i = 0; i < v[x].size(); i++) {
    t = v[x][i];
    if (t == fa[x]) continue;
    ret = ((ret + ((((num[t][k] - num[t][k ^ 1]) % 1000000007) * (n - s[t])) %
                   1000000007)) %
           1000000007);
  }
  {
    ret =
        ((ret + (((((num[1][k] - num[x][k]) - (num[1][k ^ 1] - num[x][k ^ 1])) %
                   1000000007) *
                  s[x]) %
                 1000000007)) %
         1000000007);
    ret = ((ret + n) % 1000000007);
  }
  ret = ((((ret) % 1000000007) * ((a[x]) % 1000000007)) % 1000000007);
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  for (int i = 1; i < n; i++) {
    int ta, tb;
    scanf("%d%d", &ta, &tb);
    v[ta].push_back(tb);
    v[tb].push_back(ta);
  }
  memset(vis, 0, sizeof(vis));
  memset(num, 0, sizeof(num));
  id[1] = 1;
  fa[1] = 0;
  Get_id(1);
  memset(vis, 0, sizeof(vis));
  dfs(1);
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = ((ans + cla(i)) % 1000000007);
  }
  printf("%lld\n", ((ans + 1000000007) % 1000000007));
  return 0;
}

```