#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr ll mod = 1e9 + 7, maxn = 2e5 + 100;
ll vals[maxn] = {}, d_e[maxn] = {}, d_o[maxn] = {}, u_e[maxn] = {},
   u_o[maxn] = {};
set<int> v[maxn] = {};
void dfs1(int cur, int prev) {
  d_e[cur] = 0;
  d_o[cur] = 0;
  for (auto next : v[cur]) {
    if (next == prev) continue;
    dfs1(next, cur);
    d_e[cur] += d_o[next];
    d_o[cur] += d_e[next] + 1;
  }
}
void dfs2(int cur, int prev) {
  for (auto next : v[cur]) {
    if (next == prev) continue;
    u_e[next] += u_o[cur] + d_o[cur] - d_e[next] - 1;
    u_o[next] += u_e[cur] + d_e[cur] - d_o[next] + 1;
    dfs2(next, cur);
  }
}
ll rez = 0;
void dfs3(int cur, int prev) {
  ll ret = 0;
  ret += d_e[cur] * d_e[cur] - d_o[cur] * d_o[cur];
  for (auto next : v[cur]) {
    if (next == prev) continue;
    ret += (d_e[next] + 1) * (d_e[next] + 1) - d_o[next] * d_o[next];
    dfs3(next, cur);
  }
  ret += 2ll * (d_e[cur] * u_e[cur] - d_o[cur] * u_o[cur]);
  ret += 2ll * (d_e[cur] + u_e[cur]);
  ret += 1;
  ret %= mod;
  ret += mod;
  ret %= mod;
  rez += (vals[cur] + mod) * ret;
  rez %= mod;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> vals[i];
  for (int i = 0, x, y; i < n - 1; ++i) {
    cin >> x >> y;
    v[x].insert(y);
    v[y].insert(x);
  }
  dfs1(1, -1);
  dfs2(1, -1);
  dfs3(1, -1);
  cout << rez << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
using ll = long long;
constexpr ll mod = 1e9 + 7, maxn = 2e5 + 100;
ll vals[maxn] = {}, d_e[maxn] = {}, d_o[maxn] = {}, u_e[maxn] = {},
   u_o[maxn] = {};
set<int> v[maxn] = {};
void dfs1(int cur, int prev) {
  d_e[cur] = 0;
  d_o[cur] = 0;
  for (auto next : v[cur]) {
    if (next == prev) continue;
    dfs1(next, cur);
    d_e[cur] += d_o[next];
    d_o[cur] += d_e[next] + 1;
  }
}
void dfs2(int cur, int prev) {
  for (auto next : v[cur]) {
    if (next == prev) continue;
    u_e[next] += u_o[cur] + d_o[cur] - d_e[next] - 1;
    u_o[next] += u_e[cur] + d_e[cur] - d_o[next] + 1;
    dfs2(next, cur);
  }
}
ll rez = 0;
void dfs3(int cur, int prev) {
  ll ret = 0;
  ret += d_e[cur] * d_e[cur] - d_o[cur] * d_o[cur];
  for (auto next : v[cur]) {
    if (next == prev) continue;
    ret += (d_e[next] + 1) * (d_e[next] + 1) - d_o[next] * d_o[next];
    dfs3(next, cur);
  }
  ret += 2ll * (d_e[cur] * u_e[cur] - d_o[cur] * u_o[cur]);
  ret += 2ll * (d_e[cur] + u_e[cur]);
  ret += 1;
  ret %= mod;
  ret += mod;
  ret %= mod;
  rez += (vals[cur] + mod) * ret;
  rez %= mod;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> vals[i];
  for (int i = 0, x, y; i < n - 1; ++i) {
    cin >> x >> y;
    v[x].insert(y);
    v[y].insert(x);
  }
  dfs1(1, -1);
  dfs2(1, -1);
  dfs3(1, -1);
  cout << rez << endl;
  return 0;
}

```