#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& s, vector<T> const& v) {
  s << '{';
  for (int i = 0; i < int(v.size()); i++) s << (i ? "," : "") << v[i];
  return s << '}';
}
template <typename S, typename T>
ostream& operator<<(ostream& s, pair<S, T> const& p) {
  return s << '(' << p.first << ',' << p.second << ')';
}
struct Vtx {
  int V;
  long long alt, sOdd, nOdd, sEven, nEven;
} vtxs[200000];
int N;
vector<vector<int> > adj;
constexpr long long MOD = 1000000007;
inline void addMod(long long& a, long long b) {
  a = ((a % MOD + b % MOD) % MOD + MOD) % MOD;
}
void dfs(int n, int p) {
  auto& v = vtxs[n];
  addMod(v.sOdd, v.V);
  addMod(v.nOdd, 1);
  addMod(v.alt, v.V);
  long long tnEven = 0, tnOdd = 0;
  for (int m : adj[n])
    if (m != p) {
      dfs(m, n);
      auto& v2 = vtxs[m];
      addMod(v.nOdd, v2.nEven);
      addMod(v.sOdd, v2.sEven);
      addMod(v.sOdd, v2.nEven * v.V);
      addMod(v.nEven, v2.nOdd);
      addMod(v.sEven, v2.sOdd);
      addMod(v.sEven, -v2.nOdd * v.V);
      addMod(v.alt, v2.alt);
      addMod(tnEven, v2.nEven);
      addMod(tnOdd, v2.nOdd);
    }
  long long oddPaths = 0, evenPaths = 0;
  for (int m : adj[n])
    if (m != p) {
      auto& v2 = vtxs[m];
      long long otherOdd = (tnOdd - v2.nOdd) % MOD;
      addMod(v.alt, 2 * v2.sOdd * otherOdd);
      addMod(oddPaths, v2.nOdd * otherOdd);
      long long otherEven = (tnEven - v2.nEven) % MOD;
      addMod(v.alt, 2 * v2.sEven * (otherEven + 1));
      addMod(evenPaths, v2.nEven * (otherEven + 2));
    }
  addMod(v.alt, -v.V * oddPaths);
  addMod(v.alt, v.V * evenPaths);
}
int main() {
  scanf("%d", &N);
  adj.resize(N);
  for (int n = 0; n < int(N); n++) scanf("%d", &vtxs[n].V);
  for (int a = 0; a < int(N - 1); a++) {
    int u, v;
    scanf("%d%d", &u, &v);
    --u;
    --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(0, -1);
  printf("%d\n", int(vtxs[0].alt));
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
template <typename T>
ostream& operator<<(ostream& s, vector<T> const& v) {
  s << '{';
  for (int i = 0; i < int(v.size()); i++) s << (i ? "," : "") << v[i];
  return s << '}';
}
template <typename S, typename T>
ostream& operator<<(ostream& s, pair<S, T> const& p) {
  return s << '(' << p.first << ',' << p.second << ')';
}
struct Vtx {
  int V;
  long long alt, sOdd, nOdd, sEven, nEven;
} vtxs[200000];
int N;
vector<vector<int> > adj;
constexpr long long MOD = 1000000007;
inline void addMod(long long& a, long long b) {
  a = ((a % MOD + b % MOD) % MOD + MOD) % MOD;
}
void dfs(int n, int p) {
  auto& v = vtxs[n];
  addMod(v.sOdd, v.V);
  addMod(v.nOdd, 1);
  addMod(v.alt, v.V);
  long long tnEven = 0, tnOdd = 0;
  for (int m : adj[n])
    if (m != p) {
      dfs(m, n);
      auto& v2 = vtxs[m];
      addMod(v.nOdd, v2.nEven);
      addMod(v.sOdd, v2.sEven);
      addMod(v.sOdd, v2.nEven * v.V);
      addMod(v.nEven, v2.nOdd);
      addMod(v.sEven, v2.sOdd);
      addMod(v.sEven, -v2.nOdd * v.V);
      addMod(v.alt, v2.alt);
      addMod(tnEven, v2.nEven);
      addMod(tnOdd, v2.nOdd);
    }
  long long oddPaths = 0, evenPaths = 0;
  for (int m : adj[n])
    if (m != p) {
      auto& v2 = vtxs[m];
      long long otherOdd = (tnOdd - v2.nOdd) % MOD;
      addMod(v.alt, 2 * v2.sOdd * otherOdd);
      addMod(oddPaths, v2.nOdd * otherOdd);
      long long otherEven = (tnEven - v2.nEven) % MOD;
      addMod(v.alt, 2 * v2.sEven * (otherEven + 1));
      addMod(evenPaths, v2.nEven * (otherEven + 2));
    }
  addMod(v.alt, -v.V * oddPaths);
  addMod(v.alt, v.V * evenPaths);
}
int main() {
  scanf("%d", &N);
  adj.resize(N);
  for (int n = 0; n < int(N); n++) scanf("%d", &vtxs[n].V);
  for (int a = 0; a < int(N - 1); a++) {
    int u, v;
    scanf("%d%d", &u, &v);
    --u;
    --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(0, -1);
  printf("%d\n", int(vtxs[0].alt));
}

```