#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const long long N = 2e5 + 10;
const long long mod = 1e9 + 7;
int read() {
  int a;
  scanf("%d", &a);
  return a;
}
int siz[N], n, A[N], t[N], fas[N], dep[N];
long long ans, suf[N][2], sup[N][2];
vector<int> G[N];
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  fas[u] = fa;
  siz[u] = 1;
  suf[u][0] = 1;
  for (int siz1 = G[u].size(), i = 0; i < siz1; i++) {
    int v = G[u][i];
    if (v == fa) continue;
    dfs(v, u);
    siz[u] += siz[v];
    suf[u][0] += suf[v][1];
    suf[u][1] += suf[v][0];
  }
  ans = (ans + mod +
         1ll * (suf[u][0] - suf[u][1]) * A[u] % mod * (n - siz[u] + 1) % mod) %
        mod;
  ans = (ans + mod + 1ll * A[u] * (siz[u] - 1) % mod) % mod;
  for (int siz1 = G[u].size(), i = 0; i < siz1; i++) {
    int v = G[u][i];
    if (v == fa) continue;
    ans = (ans + mod +
           1ll * (suf[v][1] - suf[v][0]) * A[u] % mod * (siz[u] - siz[v] - 1) %
               mod) %
          mod;
  }
}
bool cmp(int a, int b) { return dep[a] < dep[b]; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    A[i] = read();
    t[i] = i;
  }
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  sort(t + 1, t + 1 + n, cmp);
  for (int i = 2; i <= n; i++) {
    int u = t[i];
    sup[u][0] += sup[fas[u]][1] + suf[fas[u]][1] - suf[u][0];
    sup[u][1] += sup[fas[u]][0] + suf[fas[u]][0] - suf[u][1];
    ans =
        (mod + ans + (sup[u][0] - sup[u][1]) * siz[u] % mod * A[u] % mod) % mod;
  }
  cout << ans % mod;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
#pragma GCC optimize(2)
using namespace std;
const long long N = 2e5 + 10;
const long long mod = 1e9 + 7;
int read() {
  int a;
  scanf("%d", &a);
  return a;
}
int siz[N], n, A[N], t[N], fas[N], dep[N];
long long ans, suf[N][2], sup[N][2];
vector<int> G[N];
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  fas[u] = fa;
  siz[u] = 1;
  suf[u][0] = 1;
  for (int siz1 = G[u].size(), i = 0; i < siz1; i++) {
    int v = G[u][i];
    if (v == fa) continue;
    dfs(v, u);
    siz[u] += siz[v];
    suf[u][0] += suf[v][1];
    suf[u][1] += suf[v][0];
  }
  ans = (ans + mod +
         1ll * (suf[u][0] - suf[u][1]) * A[u] % mod * (n - siz[u] + 1) % mod) %
        mod;
  ans = (ans + mod + 1ll * A[u] * (siz[u] - 1) % mod) % mod;
  for (int siz1 = G[u].size(), i = 0; i < siz1; i++) {
    int v = G[u][i];
    if (v == fa) continue;
    ans = (ans + mod +
           1ll * (suf[v][1] - suf[v][0]) * A[u] % mod * (siz[u] - siz[v] - 1) %
               mod) %
          mod;
  }
}
bool cmp(int a, int b) { return dep[a] < dep[b]; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    A[i] = read();
    t[i] = i;
  }
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  sort(t + 1, t + 1 + n, cmp);
  for (int i = 2; i <= n; i++) {
    int u = t[i];
    sup[u][0] += sup[fas[u]][1] + suf[fas[u]][1] - suf[u][0];
    sup[u][1] += sup[fas[u]][0] + suf[fas[u]][0] - suf[u][1];
    ans =
        (mod + ans + (sup[u][0] - sup[u][1]) * siz[u] % mod * A[u] % mod) % mod;
  }
  cout << ans % mod;
}

```