#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 5, mod = 1e9 + 7;
long long sz[maxn], e[maxn], o[maxn], O[maxn], E[maxn], a[maxn], ans;
vector<long long> adj[maxn];
void getsz(long long v, long long p = -1) {
  sz[v] = 1;
  e[v] = 1;
  for (auto u : adj[v]) {
    if (u != p) {
      getsz(u, v);
      e[v] += o[u];
      o[v] += e[u];
      sz[v] += sz[u];
    }
  }
}
void getzs(long long v, long long p = -1) {
  long long ee = 0, oo = 0;
  for (auto u : adj[v]) {
    if (u != p) {
      ee += e[u];
      oo += o[u];
      ans -= a[v] * e[u] * (sz[v] - 1 - sz[u]);
      ans += a[v] * o[u] * (sz[v] - 1 - sz[u]);
      ans %= mod;
    }
  }
  for (auto u : adj[v]) {
    if (u != p) {
      E[u] = O[v] + ee - e[u];
      O[u] = 1 + E[v] + oo - o[u];
      getzs(u, v);
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n - 1; i++) {
    long long v, u;
    cin >> v >> u;
    v--;
    u--;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  getsz(0);
  getzs(0);
  for (long long i = 0; i < n; i++) {
    ans += a[i] * (e[i] * (n - sz[i] + 1) + (E[i] + 1) * sz[i] - 1);
    ans -= a[i] * (o[i] * (n - sz[i] + 1) + O[i] * sz[i]);
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  cout << ans;
}

### 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>
using namespace std;
const long long maxn = 2e5 + 5, mod = 1e9 + 7;
long long sz[maxn], e[maxn], o[maxn], O[maxn], E[maxn], a[maxn], ans;
vector<long long> adj[maxn];
void getsz(long long v, long long p = -1) {
  sz[v] = 1;
  e[v] = 1;
  for (auto u : adj[v]) {
    if (u != p) {
      getsz(u, v);
      e[v] += o[u];
      o[v] += e[u];
      sz[v] += sz[u];
    }
  }
}
void getzs(long long v, long long p = -1) {
  long long ee = 0, oo = 0;
  for (auto u : adj[v]) {
    if (u != p) {
      ee += e[u];
      oo += o[u];
      ans -= a[v] * e[u] * (sz[v] - 1 - sz[u]);
      ans += a[v] * o[u] * (sz[v] - 1 - sz[u]);
      ans %= mod;
    }
  }
  for (auto u : adj[v]) {
    if (u != p) {
      E[u] = O[v] + ee - e[u];
      O[u] = 1 + E[v] + oo - o[u];
      getzs(u, v);
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n - 1; i++) {
    long long v, u;
    cin >> v >> u;
    v--;
    u--;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  getsz(0);
  getzs(0);
  for (long long i = 0; i < n; i++) {
    ans += a[i] * (e[i] * (n - sz[i] + 1) + (E[i] + 1) * sz[i] - 1);
    ans -= a[i] * (o[i] * (n - sz[i] + 1) + O[i] * sz[i]);
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  cout << ans;
}

```