#include <bits/stdc++.h>
using namespace std;
const long double PI = atan(1.0) * 4;
const int64_t INF = 8e18;
const int32_t LOG = 21;
const long long int MOD = 1e9 + 7;
long long int a[200005];
long long int siz[200005];
long long int cnt[200005][2];
long long int dp[200005][2];
vector<long long int> v[200005];
long long int ans = 0;
long long int n;
void dfs(long long int i, long long int prev) {
  cnt[i][1] = 1;
  siz[i]++;
  for (auto it : v[i]) {
    if (it != prev) {
      dfs(it, i);
      siz[i] += siz[it];
      cnt[i][0] += cnt[it][1];
      cnt[i][1] += cnt[it][0];
    }
  }
}
void dfs1(long long int i, long long int prev) {
  dp[i][1] = 1;
  long long int toAdd = 0;
  for (auto it : v[i]) {
    if (it != prev) {
      dp[i][0] += cnt[it][1];
      dp[i][1] += cnt[it][0];
      dp[i][0] %= MOD;
      dp[i][1] %= MOD;
      toAdd += (((cnt[it][0] - cnt[it][1] + MOD) * (n - siz[it])) % MOD) *
               (a[i] + MOD);
      toAdd %= MOD;
    }
  }
  if (prev != -1) {
    toAdd += (((dp[prev][0] - cnt[i][1] - dp[prev][1] + cnt[i][0] + 2 * MOD) *
               (siz[i])) %
              MOD) *
             (a[i] + MOD);
    dp[i][0] += (dp[prev][1] - cnt[i][0] + MOD);
    dp[i][1] += (dp[prev][0] - cnt[i][1] + MOD);
    dp[i][0] %= MOD;
    dp[i][1] %= MOD;
    toAdd %= MOD;
  }
  toAdd += n * (a[i] + MOD);
  toAdd %= MOD;
  ans += toAdd;
  while (ans >= MOD) ans -= MOD;
  for (auto it : v[i]) {
    if (it != prev) dfs1(it, i);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int tests = 1;
  for (long long int ii = 0; ii < tests; ii++) {
    cin >> n;
    for (long long int i = 1; i < n + 1; i++) cin >> a[i];
    for (long long int i = 1; i < n; i++) {
      long long int x, y;
      cin >> x >> y;
      v[x].push_back(y);
      v[y].push_back(x);
    }
  }
  dfs(1, -1);
  dfs1(1, -1);
  cout << ans << '\n';
  return 0;
}

### Prompt

Generate 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 double PI = atan(1.0) * 4;
const int64_t INF = 8e18;
const int32_t LOG = 21;
const long long int MOD = 1e9 + 7;
long long int a[200005];
long long int siz[200005];
long long int cnt[200005][2];
long long int dp[200005][2];
vector<long long int> v[200005];
long long int ans = 0;
long long int n;
void dfs(long long int i, long long int prev) {
  cnt[i][1] = 1;
  siz[i]++;
  for (auto it : v[i]) {
    if (it != prev) {
      dfs(it, i);
      siz[i] += siz[it];
      cnt[i][0] += cnt[it][1];
      cnt[i][1] += cnt[it][0];
    }
  }
}
void dfs1(long long int i, long long int prev) {
  dp[i][1] = 1;
  long long int toAdd = 0;
  for (auto it : v[i]) {
    if (it != prev) {
      dp[i][0] += cnt[it][1];
      dp[i][1] += cnt[it][0];
      dp[i][0] %= MOD;
      dp[i][1] %= MOD;
      toAdd += (((cnt[it][0] - cnt[it][1] + MOD) * (n - siz[it])) % MOD) *
               (a[i] + MOD);
      toAdd %= MOD;
    }
  }
  if (prev != -1) {
    toAdd += (((dp[prev][0] - cnt[i][1] - dp[prev][1] + cnt[i][0] + 2 * MOD) *
               (siz[i])) %
              MOD) *
             (a[i] + MOD);
    dp[i][0] += (dp[prev][1] - cnt[i][0] + MOD);
    dp[i][1] += (dp[prev][0] - cnt[i][1] + MOD);
    dp[i][0] %= MOD;
    dp[i][1] %= MOD;
    toAdd %= MOD;
  }
  toAdd += n * (a[i] + MOD);
  toAdd %= MOD;
  ans += toAdd;
  while (ans >= MOD) ans -= MOD;
  for (auto it : v[i]) {
    if (it != prev) dfs1(it, i);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int tests = 1;
  for (long long int ii = 0; ii < tests; ii++) {
    cin >> n;
    for (long long int i = 1; i < n + 1; i++) cin >> a[i];
    for (long long int i = 1; i < n; i++) {
      long long int x, y;
      cin >> x >> y;
      v[x].push_back(y);
      v[y].push_back(x);
    }
  }
  dfs(1, -1);
  dfs1(1, -1);
  cout << ans << '\n';
  return 0;
}

```