#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long N = 2e5 + 10;
long long n;
long long a[N];
vector<long long> g[N];
long long col[N];
long long dp[N][2];
long long res = 0;
void dfs(long long node, long long par = -1, long long cc = 0) {
  col[node] = cc;
  dp[node][0] = 1;
  for (auto x : g[node])
    if (x != par) {
      dfs(x, node, cc ^ 1);
      dp[node][1] += dp[x][0];
      dp[node][0] += dp[x][1];
    }
}
void dfs2(long long node, long long par = -1) {
  long long temp = 0;
  for (auto x : g[node]) {
    if (x == par) continue;
    dfs2(x, node);
    temp += dp[x][1] * (n - dp[x][0] - dp[x][1]);
    temp -= dp[x][0] * (n - dp[x][0] - dp[x][1]);
    temp %= MOD;
    if (temp < 0) temp += MOD;
  }
  if (col[1] == col[node]) {
    temp += (dp[1][0] - dp[node][0]) * (dp[node][0] + dp[node][1]);
    temp -= (dp[1][1] - dp[node][1]) * (dp[node][0] + dp[node][1]);
  } else {
    temp += (dp[1][1] - dp[node][0]) * (dp[node][0] + dp[node][1]);
    temp -= (dp[1][0] - dp[node][1]) * (dp[node][0] + dp[node][1]);
  }
  temp += n;
  temp %= MOD;
  if (temp < 0) temp += MOD;
  temp *= a[node];
  temp %= MOD;
  if (temp < 0) temp += MOD;
  res += temp;
  res %= MOD;
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i + 1];
  long long u, v;
  for (long long i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  dfs2(1);
  cout << res;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

### Prompt

Create a solution in Cpp for 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 long long MOD = 1e9 + 7;
const long long N = 2e5 + 10;
long long n;
long long a[N];
vector<long long> g[N];
long long col[N];
long long dp[N][2];
long long res = 0;
void dfs(long long node, long long par = -1, long long cc = 0) {
  col[node] = cc;
  dp[node][0] = 1;
  for (auto x : g[node])
    if (x != par) {
      dfs(x, node, cc ^ 1);
      dp[node][1] += dp[x][0];
      dp[node][0] += dp[x][1];
    }
}
void dfs2(long long node, long long par = -1) {
  long long temp = 0;
  for (auto x : g[node]) {
    if (x == par) continue;
    dfs2(x, node);
    temp += dp[x][1] * (n - dp[x][0] - dp[x][1]);
    temp -= dp[x][0] * (n - dp[x][0] - dp[x][1]);
    temp %= MOD;
    if (temp < 0) temp += MOD;
  }
  if (col[1] == col[node]) {
    temp += (dp[1][0] - dp[node][0]) * (dp[node][0] + dp[node][1]);
    temp -= (dp[1][1] - dp[node][1]) * (dp[node][0] + dp[node][1]);
  } else {
    temp += (dp[1][1] - dp[node][0]) * (dp[node][0] + dp[node][1]);
    temp -= (dp[1][0] - dp[node][1]) * (dp[node][0] + dp[node][1]);
  }
  temp += n;
  temp %= MOD;
  if (temp < 0) temp += MOD;
  temp *= a[node];
  temp %= MOD;
  if (temp < 0) temp += MOD;
  res += temp;
  res %= MOD;
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i + 1];
  long long u, v;
  for (long long i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  dfs2(1);
  cout << res;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

```