#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000000 + 7;
struct Node {
  long long vp, vn;
  vector<int> to;
  int down0;
  int down1;
  int up0;
  int up1;
};
vector<Node> g;
long long norm(long long a) {
  while (a < 0) a += MOD;
  a %= MOD;
  return a;
}
void Dfs1(int i, int from) {
  g[i].down0 = 0;
  g[i].down1 = 1;
  for (int j = 0; j < int(g[i].to.size()); ++j) {
    int to = g[i].to[j];
    if (to == from) continue;
    Dfs1(to, i);
    g[i].down0 += g[to].down1;
    g[i].down1 += g[to].down0;
  }
}
void Dfs2(int i, int from, int back0, int back1) {
  g[i].up0 = back0;
  g[i].up1 = back1;
  int up0 = back1;
  int up1 = back0 + 1;
  for (int j = 0; j < int(g[i].to.size()); ++j) {
    int to = g[i].to[j];
    if (to == from) continue;
    up0 += g[to].down1;
    up1 += g[to].down0;
  }
  for (int j = 0; j < int(g[i].to.size()); ++j) {
    int to = g[i].to[j];
    if (to == from) continue;
    Dfs2(to, i, up0 - g[to].down1, up1 - g[to].down0);
  }
}
long long Dfs3(int i, int from) {
  int c0 = 0;
  int c1 = 0;
  for (int j = 0; j < int(g[i].to.size()); ++j) {
    int to = g[i].to[j];
    if (to == from) {
      c0 += g[i].up0;
      c1 += g[i].up1;
    } else {
      c0 += g[to].down0;
      c1 += g[to].down1;
    }
  }
  long long total = (c0 + c1 + 1) * g[i].vp % MOD;
  for (int j = 0; j < int(g[i].to.size()); ++j) {
    int to = g[i].to[j];
    int cq0, cq1;
    if (to == from) {
      cq0 = g[i].up0;
      cq1 = g[i].up1;
    } else {
      cq0 = g[to].down0;
      cq1 = g[to].down1;
    }
    int cc0 = c0 - cq0;
    int cc1 = c1 - cq1;
    int cc = cc0 + cc1 + 1;
    total += ((long long)cq0 * cc) % MOD * g[i].vp;
    total += ((long long)cq1 * cc) % MOD * g[i].vn;
    total %= MOD;
  }
  for (int j = 0; j < int(g[i].to.size()); ++j) {
    int to = g[i].to[j];
    if (to == from) continue;
    total += Dfs3(to, i);
  }
  return total % MOD;
}
int main() {
  int n;
  cin >> n;
  g.resize(n);
  for (int i = 0; i < int(n); ++i) {
    long long a;
    cin >> a;
    g[i].vp = norm(a);
    g[i].vn = norm(-a);
  }
  for (int i = 0; i < int(n - 1); ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    g[a].to.push_back(b);
    g[b].to.push_back(a);
  }
  Dfs1(0, -1);
  Dfs2(0, -1, 0, 0);
  long long res = Dfs3(0, -1);
  cout << res << endl;
  return 0;
}

### Prompt

Generate 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 long long MOD = 1000000000 + 7;
struct Node {
  long long vp, vn;
  vector<int> to;
  int down0;
  int down1;
  int up0;
  int up1;
};
vector<Node> g;
long long norm(long long a) {
  while (a < 0) a += MOD;
  a %= MOD;
  return a;
}
void Dfs1(int i, int from) {
  g[i].down0 = 0;
  g[i].down1 = 1;
  for (int j = 0; j < int(g[i].to.size()); ++j) {
    int to = g[i].to[j];
    if (to == from) continue;
    Dfs1(to, i);
    g[i].down0 += g[to].down1;
    g[i].down1 += g[to].down0;
  }
}
void Dfs2(int i, int from, int back0, int back1) {
  g[i].up0 = back0;
  g[i].up1 = back1;
  int up0 = back1;
  int up1 = back0 + 1;
  for (int j = 0; j < int(g[i].to.size()); ++j) {
    int to = g[i].to[j];
    if (to == from) continue;
    up0 += g[to].down1;
    up1 += g[to].down0;
  }
  for (int j = 0; j < int(g[i].to.size()); ++j) {
    int to = g[i].to[j];
    if (to == from) continue;
    Dfs2(to, i, up0 - g[to].down1, up1 - g[to].down0);
  }
}
long long Dfs3(int i, int from) {
  int c0 = 0;
  int c1 = 0;
  for (int j = 0; j < int(g[i].to.size()); ++j) {
    int to = g[i].to[j];
    if (to == from) {
      c0 += g[i].up0;
      c1 += g[i].up1;
    } else {
      c0 += g[to].down0;
      c1 += g[to].down1;
    }
  }
  long long total = (c0 + c1 + 1) * g[i].vp % MOD;
  for (int j = 0; j < int(g[i].to.size()); ++j) {
    int to = g[i].to[j];
    int cq0, cq1;
    if (to == from) {
      cq0 = g[i].up0;
      cq1 = g[i].up1;
    } else {
      cq0 = g[to].down0;
      cq1 = g[to].down1;
    }
    int cc0 = c0 - cq0;
    int cc1 = c1 - cq1;
    int cc = cc0 + cc1 + 1;
    total += ((long long)cq0 * cc) % MOD * g[i].vp;
    total += ((long long)cq1 * cc) % MOD * g[i].vn;
    total %= MOD;
  }
  for (int j = 0; j < int(g[i].to.size()); ++j) {
    int to = g[i].to[j];
    if (to == from) continue;
    total += Dfs3(to, i);
  }
  return total % MOD;
}
int main() {
  int n;
  cin >> n;
  g.resize(n);
  for (int i = 0; i < int(n); ++i) {
    long long a;
    cin >> a;
    g[i].vp = norm(a);
    g[i].vn = norm(-a);
  }
  for (int i = 0; i < int(n - 1); ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    g[a].to.push_back(b);
    g[b].to.push_back(a);
  }
  Dfs1(0, -1);
  Dfs2(0, -1, 0, 0);
  long long res = Dfs3(0, -1);
  cout << res << endl;
  return 0;
}

```