#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const long long N = 2e5 + 5;
long long n, v[N], ans, num[N];
vector<long long> g[N];
long long dp[N][2];
void add(long long &a, long long b, long long c, long long d) {
  a += (b * c + d);
  a %= 1000000007;
  a += 1000000007;
  a %= 1000000007;
}
void dfs(long long at, long long fr = -1) {
  num[at] = 1;
  dp[at][0] = v[at];
  dp[at][1] = -v[at];
  for (long long to : g[at]) {
    if (to == fr) continue;
    dfs(to, at);
    add(dp[at][0], num[to], v[at], dp[to][1]);
    add(dp[at][1], num[to], -v[at], dp[to][0]);
    num[at] += num[to];
  }
}
void update(long long at, long long fr = -1) {
  ans = (ans + dp[at][0]) % 1000000007;
  ans = (ans + 1000000007) % 1000000007;
  for (long long to : g[at]) {
    if (to == fr) continue;
    long long pre0 = dp[at][0] - (num[to] * v[at] + dp[to][1]);
    long long pre1 = dp[at][1] - (num[to] * (-v[at]) + dp[to][0]);
    pre0 %= 1000000007, pre1 %= 1000000007;
    pre0 = (pre0 + 1000000007) % 1000000007,
    pre1 = (pre1 + 1000000007) % 1000000007;
    long long f = num[to];
    num[at] -= num[to];
    num[to] += num[at];
    add(dp[to][0], num[at], v[to], pre1);
    add(dp[to][1], num[at], -v[to], pre0);
    num[at] += f;
    update(to, at);
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (long long i = 1, u, v; i < n; i++) {
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  update(0);
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 INF = 1e18;
const long long N = 2e5 + 5;
long long n, v[N], ans, num[N];
vector<long long> g[N];
long long dp[N][2];
void add(long long &a, long long b, long long c, long long d) {
  a += (b * c + d);
  a %= 1000000007;
  a += 1000000007;
  a %= 1000000007;
}
void dfs(long long at, long long fr = -1) {
  num[at] = 1;
  dp[at][0] = v[at];
  dp[at][1] = -v[at];
  for (long long to : g[at]) {
    if (to == fr) continue;
    dfs(to, at);
    add(dp[at][0], num[to], v[at], dp[to][1]);
    add(dp[at][1], num[to], -v[at], dp[to][0]);
    num[at] += num[to];
  }
}
void update(long long at, long long fr = -1) {
  ans = (ans + dp[at][0]) % 1000000007;
  ans = (ans + 1000000007) % 1000000007;
  for (long long to : g[at]) {
    if (to == fr) continue;
    long long pre0 = dp[at][0] - (num[to] * v[at] + dp[to][1]);
    long long pre1 = dp[at][1] - (num[to] * (-v[at]) + dp[to][0]);
    pre0 %= 1000000007, pre1 %= 1000000007;
    pre0 = (pre0 + 1000000007) % 1000000007,
    pre1 = (pre1 + 1000000007) % 1000000007;
    long long f = num[to];
    num[at] -= num[to];
    num[to] += num[at];
    add(dp[to][0], num[at], v[to], pre1);
    add(dp[to][1], num[at], -v[to], pre0);
    num[at] += f;
    update(to, at);
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (long long i = 1, u, v; i < n; i++) {
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  update(0);
  cout << ans << endl;
  return 0;
}

```