#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 200005;
vector<int> val;
int cmp[MAX_N];
vector<int> G[MAX_N];
int cnt[MAX_N][2];
int al[2];
int ans;
long long add(long long x, long long y) { return (x + y) % 1000000007; }
long long sub(long long x, long long y) {
  return (x + 1000000007 - y) % 1000000007;
}
long long mul(long long x, long long y) { return x * y % 1000000007; }
void color(int u, int p, int kind) {
  cmp[u] = kind;
  al[kind]++;
  for (auto&(v) : (G[u])) {
    if (v != p) {
      color(v, u, kind ^ 1);
    }
  }
}
void dfs(int u, int p) {
  cnt[u][cmp[u]] = 1;
  int sm = 0;
  for (auto&(v) : (G[u])) {
    if (v != p) {
      dfs(v, u);
      for (int k = 0; k < (int)(2); ++k) {
        cnt[u][k] += cnt[v][k];
        sm += cnt[v][k];
      }
    }
  }
  for (auto&(v) : (G[u])) {
    if (v != p) {
      ans = add(ans, mul(mul(sub(cnt[v][cmp[u]], cnt[v][cmp[u] ^ 1]),
                             sm - cnt[v][0] - cnt[v][1]),
                         val[u]));
      ans = add(ans, mul(sub(cnt[v][cmp[u]], cnt[v][cmp[u] ^ 1]), val[u]));
      ans = add(ans, mul(sub(cnt[v][cmp[u]], cnt[v][cmp[u] ^ 1]),
                         mul(val[u], al[cmp[u]] - cnt[u][cmp[u]] +
                                         al[cmp[u] ^ 1] - cnt[u][cmp[u] ^ 1])));
    }
  }
  ans = add(ans, mul(sub(al[cmp[u]] - cnt[u][cmp[u]],
                         al[cmp[u] ^ 1] - cnt[u][cmp[u] ^ 1]),
                     val[u]));
  ans = add(ans, mul(sub(al[cmp[u]] - cnt[u][cmp[u]],
                         al[cmp[u] ^ 1] - cnt[u][cmp[u] ^ 1]),
                     mul(val[u], cnt[u][0] + cnt[u][1] - 1)));
  ans = add(ans, mul(al[cmp[u]] - cnt[u][cmp[u]] + al[cmp[u] ^ 1] -
                         cnt[u][cmp[u] ^ 1],
                     val[u]));
  ans = add(ans, mul(cnt[u][0] + cnt[u][1], val[u]));
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  val.resize(n);
  for (int i = 0; i < (int)(n); ++i) {
    cin >> val[i];
    val[i] = (val[i] + 1000000007) % 1000000007;
  }
  for (int i = 0; i < (int)(n - 1); ++i) {
    int a, b;
    cin >> a >> b;
    G[a - 1].push_back(b - 1), G[b - 1].push_back(a - 1);
  }
  color(0, -1, 0);
  dfs(0, -1);
  cout << ans << "\n";
  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;
const int MAX_N = 200005;
vector<int> val;
int cmp[MAX_N];
vector<int> G[MAX_N];
int cnt[MAX_N][2];
int al[2];
int ans;
long long add(long long x, long long y) { return (x + y) % 1000000007; }
long long sub(long long x, long long y) {
  return (x + 1000000007 - y) % 1000000007;
}
long long mul(long long x, long long y) { return x * y % 1000000007; }
void color(int u, int p, int kind) {
  cmp[u] = kind;
  al[kind]++;
  for (auto&(v) : (G[u])) {
    if (v != p) {
      color(v, u, kind ^ 1);
    }
  }
}
void dfs(int u, int p) {
  cnt[u][cmp[u]] = 1;
  int sm = 0;
  for (auto&(v) : (G[u])) {
    if (v != p) {
      dfs(v, u);
      for (int k = 0; k < (int)(2); ++k) {
        cnt[u][k] += cnt[v][k];
        sm += cnt[v][k];
      }
    }
  }
  for (auto&(v) : (G[u])) {
    if (v != p) {
      ans = add(ans, mul(mul(sub(cnt[v][cmp[u]], cnt[v][cmp[u] ^ 1]),
                             sm - cnt[v][0] - cnt[v][1]),
                         val[u]));
      ans = add(ans, mul(sub(cnt[v][cmp[u]], cnt[v][cmp[u] ^ 1]), val[u]));
      ans = add(ans, mul(sub(cnt[v][cmp[u]], cnt[v][cmp[u] ^ 1]),
                         mul(val[u], al[cmp[u]] - cnt[u][cmp[u]] +
                                         al[cmp[u] ^ 1] - cnt[u][cmp[u] ^ 1])));
    }
  }
  ans = add(ans, mul(sub(al[cmp[u]] - cnt[u][cmp[u]],
                         al[cmp[u] ^ 1] - cnt[u][cmp[u] ^ 1]),
                     val[u]));
  ans = add(ans, mul(sub(al[cmp[u]] - cnt[u][cmp[u]],
                         al[cmp[u] ^ 1] - cnt[u][cmp[u] ^ 1]),
                     mul(val[u], cnt[u][0] + cnt[u][1] - 1)));
  ans = add(ans, mul(al[cmp[u]] - cnt[u][cmp[u]] + al[cmp[u] ^ 1] -
                         cnt[u][cmp[u] ^ 1],
                     val[u]));
  ans = add(ans, mul(cnt[u][0] + cnt[u][1], val[u]));
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  val.resize(n);
  for (int i = 0; i < (int)(n); ++i) {
    cin >> val[i];
    val[i] = (val[i] + 1000000007) % 1000000007;
  }
  for (int i = 0; i < (int)(n - 1); ++i) {
    int a, b;
    cin >> a >> b;
    G[a - 1].push_back(b - 1), G[b - 1].push_back(a - 1);
  }
  color(0, -1, 0);
  dfs(0, -1);
  cout << ans << "\n";
  return 0;
}

```