#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
void add(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
  if (x < 0) x += MOD;
}
int _add(int x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
  if (x < 0) x += MOD;
  return x;
}
int mul(int a, int b) { return (a * 1LL * b) % MOD; }
const int N = 200010;
int ans = 0;
int a[N];
int sz[N][2], dp[N][2];
vector<int> edges[N];
void dfs(int v, int p = -1) {
  for (int iter = 0; iter < 2; iter++) {
    sz[v][0] = 0;
    sz[v][1] = 0;
    dp[v][0] = 0;
    dp[v][1] = 0;
    int sum_odd = 0, sum_even = 0;
    for (int u : edges[v]) {
      if (u == p) continue;
      if (iter == 0) dfs(u, v);
      if (sz[u][1]) {
        add(ans, mul(dp[u][1], sz[v][0] + sz[v][1]));
        add(ans, mul(MOD - a[v], mul(sz[u][1], sz[v][0] + sz[v][1])));
        add(ans, mul(sum_odd, sz[u][1]));
        add(ans, mul(MOD - sum_even, sz[u][1]));
      }
      if (sz[u][0]) {
        add(ans, mul(dp[u][0], sz[v][0] + sz[v][1]));
        add(ans, mul(a[v], mul(sz[u][0], sz[v][0] + sz[v][1])));
        add(ans, mul(MOD - sum_odd, sz[u][0]));
        add(ans, mul(sum_even, sz[u][0]));
      }
      sz[v][0] += sz[u][1];
      sz[v][1] += sz[u][0];
      add(dp[v][0], _add(dp[u][1], mul(sz[u][1], MOD - a[v])));
      add(dp[v][1], _add(dp[u][0], mul(sz[u][0], a[v])));
      add(sum_odd, dp[u][1]);
      add(sum_even, dp[u][0]);
    }
    reverse((edges[v]).begin(), (edges[v]).end());
  }
  add(ans, mul(2, dp[v][1]));
  sz[v][1]++;
  add(dp[v][1], a[v]);
  add(ans, a[v]);
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] < 0) a[i] += MOD;
  }
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    edges[v].push_back(u);
    edges[u].push_back(v);
  }
  dfs(1);
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  solve();
}

### 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 int MOD = 1e9 + 7;
void add(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
  if (x < 0) x += MOD;
}
int _add(int x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
  if (x < 0) x += MOD;
  return x;
}
int mul(int a, int b) { return (a * 1LL * b) % MOD; }
const int N = 200010;
int ans = 0;
int a[N];
int sz[N][2], dp[N][2];
vector<int> edges[N];
void dfs(int v, int p = -1) {
  for (int iter = 0; iter < 2; iter++) {
    sz[v][0] = 0;
    sz[v][1] = 0;
    dp[v][0] = 0;
    dp[v][1] = 0;
    int sum_odd = 0, sum_even = 0;
    for (int u : edges[v]) {
      if (u == p) continue;
      if (iter == 0) dfs(u, v);
      if (sz[u][1]) {
        add(ans, mul(dp[u][1], sz[v][0] + sz[v][1]));
        add(ans, mul(MOD - a[v], mul(sz[u][1], sz[v][0] + sz[v][1])));
        add(ans, mul(sum_odd, sz[u][1]));
        add(ans, mul(MOD - sum_even, sz[u][1]));
      }
      if (sz[u][0]) {
        add(ans, mul(dp[u][0], sz[v][0] + sz[v][1]));
        add(ans, mul(a[v], mul(sz[u][0], sz[v][0] + sz[v][1])));
        add(ans, mul(MOD - sum_odd, sz[u][0]));
        add(ans, mul(sum_even, sz[u][0]));
      }
      sz[v][0] += sz[u][1];
      sz[v][1] += sz[u][0];
      add(dp[v][0], _add(dp[u][1], mul(sz[u][1], MOD - a[v])));
      add(dp[v][1], _add(dp[u][0], mul(sz[u][0], a[v])));
      add(sum_odd, dp[u][1]);
      add(sum_even, dp[u][0]);
    }
    reverse((edges[v]).begin(), (edges[v]).end());
  }
  add(ans, mul(2, dp[v][1]));
  sz[v][1]++;
  add(dp[v][1], a[v]);
  add(ans, a[v]);
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] < 0) a[i] += MOD;
  }
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    edges[v].push_back(u);
    edges[u].push_back(v);
  }
  dfs(1);
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  solve();
}

```