#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
long long add(long long a, long long b) {
  long long x = a + b;
  if (x >= 1000000007) x -= 1000000007;
  if (x < 0) x += 1000000007;
  return x;
}
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
long long pw(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return ans;
}
long long ans = 0;
int n, val[200002];
vector<int> v[200002];
int pd[200002], id[200002];
void dfs(int dad, int nod) {
  pd[nod] = 1;
  for (int i = 0; i < v[nod].size(); ++i) {
    int vecin = v[nod][i];
    if (vecin == dad) continue;
    dfs(nod, vecin);
    pd[nod] += id[vecin];
    id[nod] += pd[vecin];
  }
}
void dfs2(int dad, int nod, int pr, int imp) {
  long long ct = 0;
  ct = add(ct, mul(val[nod], mul(pr, pd[nod] + id[nod])));
  ct = add(ct, -mul(val[nod], mul(imp, pd[nod] + id[nod])));
  ct = add(ct, mul(val[nod], n));
  int sim = 0, sp = 0;
  for (int i = 0; i < v[nod].size(); ++i) {
    int vecin = v[nod][i];
    if (vecin == dad) continue;
    sim += id[vecin];
    sp += pd[vecin];
  }
  for (int i = 0; i < v[nod].size(); ++i) {
    int vecin = v[nod][i];
    if (vecin == dad) continue;
    sim -= id[vecin];
    sp -= pd[vecin];
    ct = add(ct, mul(val[nod], mul(id[vecin], pr + imp + sim + sp + 1)));
    ct = add(ct, -mul(val[nod], mul(pd[vecin], pr + imp + sim + sp + 1)));
    dfs2(nod, vecin, imp + sp, pr + sim + 1);
    sim += id[vecin];
    sp += pd[vecin];
  }
  ans = add(ans, ct);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> val[i];
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(0, 1);
  dfs2(0, 1, 0, 0);
  cout << ans;
  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>
#pragma GCC optimize("O3")
using namespace std;
long long add(long long a, long long b) {
  long long x = a + b;
  if (x >= 1000000007) x -= 1000000007;
  if (x < 0) x += 1000000007;
  return x;
}
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
long long pw(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return ans;
}
long long ans = 0;
int n, val[200002];
vector<int> v[200002];
int pd[200002], id[200002];
void dfs(int dad, int nod) {
  pd[nod] = 1;
  for (int i = 0; i < v[nod].size(); ++i) {
    int vecin = v[nod][i];
    if (vecin == dad) continue;
    dfs(nod, vecin);
    pd[nod] += id[vecin];
    id[nod] += pd[vecin];
  }
}
void dfs2(int dad, int nod, int pr, int imp) {
  long long ct = 0;
  ct = add(ct, mul(val[nod], mul(pr, pd[nod] + id[nod])));
  ct = add(ct, -mul(val[nod], mul(imp, pd[nod] + id[nod])));
  ct = add(ct, mul(val[nod], n));
  int sim = 0, sp = 0;
  for (int i = 0; i < v[nod].size(); ++i) {
    int vecin = v[nod][i];
    if (vecin == dad) continue;
    sim += id[vecin];
    sp += pd[vecin];
  }
  for (int i = 0; i < v[nod].size(); ++i) {
    int vecin = v[nod][i];
    if (vecin == dad) continue;
    sim -= id[vecin];
    sp -= pd[vecin];
    ct = add(ct, mul(val[nod], mul(id[vecin], pr + imp + sim + sp + 1)));
    ct = add(ct, -mul(val[nod], mul(pd[vecin], pr + imp + sim + sp + 1)));
    dfs2(nod, vecin, imp + sp, pr + sim + 1);
    sim += id[vecin];
    sp += pd[vecin];
  }
  ans = add(ans, ct);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> val[i];
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(0, 1);
  dfs2(0, 1, 0, 0);
  cout << ans;
  return 0;
}

```