#include <bits/stdc++.h>
#pragma GCC optimize("-Ofast", "-funroll-all-loops")
using namespace std;
const long long mod = 1e9 + 7;
const long long N = 2e5 + 10;
long long n, a[N], dp[N][2], dep[N], ans[N], sz[N], now;
vector<long long> g[N];
inline void add(long long a, long long b) {
  g[a].push_back(b), g[b].push_back(a);
}
void dfs1(long long x, long long fa) {
  dep[x] = dep[fa] + 1;
  dp[x][dep[x] % 2]++;
  sz[x] = 1;
  for (long long to : g[x])
    if (to != fa) {
      dfs1(to, x);
      sz[x] += sz[to];
      dp[x][0] += dp[to][0], dp[x][1] += dp[to][1];
      ans[x] = (ans[x] + dp[to][dep[x] % 2] * (n - sz[to]) % mod) % mod;
      ans[x] = (ans[x] - dp[to][1 ^ (dep[x] % 2)] * (n - sz[to]) % mod) % mod;
    }
}
void dfs2(long long x, long long fa) {
  for (long long to : g[x])
    if (to != fa) {
      ans[to] = (ans[to] +
                 (dp[n][dep[to] % 2] - dp[to][dep[to] % 2]) * sz[to] % mod) %
                mod;
      ans[to] =
          (ans[to] - (dp[n][1 ^ (dep[to] % 2)] - dp[to][1 ^ (dep[to] % 2)]) *
                         sz[to] % mod) %
          mod;
      dfs2(to, x);
    }
}
signed main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1, a, b; i < n; i++) scanf("%lld %lld", &a, &b), add(a, b);
  dfs1(n, 0);
  dfs2(n, 0);
  for (long long i = 1; i <= n; i++) ans[i] = (ans[i] + n) % mod;
  for (long long i = 1; i <= n; i++) now = (now + ans[i] * a[i] % mod) % mod;
  cout << (now + mod) % mod;
  return 0;
}

### Prompt

Please formulate 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("-Ofast", "-funroll-all-loops")
using namespace std;
const long long mod = 1e9 + 7;
const long long N = 2e5 + 10;
long long n, a[N], dp[N][2], dep[N], ans[N], sz[N], now;
vector<long long> g[N];
inline void add(long long a, long long b) {
  g[a].push_back(b), g[b].push_back(a);
}
void dfs1(long long x, long long fa) {
  dep[x] = dep[fa] + 1;
  dp[x][dep[x] % 2]++;
  sz[x] = 1;
  for (long long to : g[x])
    if (to != fa) {
      dfs1(to, x);
      sz[x] += sz[to];
      dp[x][0] += dp[to][0], dp[x][1] += dp[to][1];
      ans[x] = (ans[x] + dp[to][dep[x] % 2] * (n - sz[to]) % mod) % mod;
      ans[x] = (ans[x] - dp[to][1 ^ (dep[x] % 2)] * (n - sz[to]) % mod) % mod;
    }
}
void dfs2(long long x, long long fa) {
  for (long long to : g[x])
    if (to != fa) {
      ans[to] = (ans[to] +
                 (dp[n][dep[to] % 2] - dp[to][dep[to] % 2]) * sz[to] % mod) %
                mod;
      ans[to] =
          (ans[to] - (dp[n][1 ^ (dep[to] % 2)] - dp[to][1 ^ (dep[to] % 2)]) *
                         sz[to] % mod) %
          mod;
      dfs2(to, x);
    }
}
signed main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1, a, b; i < n; i++) scanf("%lld %lld", &a, &b), add(a, b);
  dfs1(n, 0);
  dfs2(n, 0);
  for (long long i = 1; i <= n; i++) ans[i] = (ans[i] + n) % mod;
  for (long long i = 1; i <= n; i++) now = (now + ans[i] * a[i] % mod) % mod;
  cout << (now + mod) % mod;
  return 0;
}

```