#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 200005;
long long ans, odd_pairs, even_pairs, odd[2], even[2], done[N], sz[N], a[N];
vector<int> adj[N];
void calc_subsize(int node, int par) {
  sz[node] = 0;
  for (int child : adj[node]) {
    if (child != par && !done[child]) {
      calc_subsize(child, node);
      sz[node] += sz[child];
    }
  }
  sz[node]++;
}
int find_centroid(int node, int par, int conditional_size) {
  for (int child : adj[node]) {
    if (child != par && !done[child] && sz[child] > conditional_size) {
      return find_centroid(child, node, conditional_size);
    }
  }
  return node;
}
void dfs(int node, int par, int even_sum, int odd_sum, int h, int typ) {
  if (h & 1) {
    odd_sum += a[node];
    odd_sum %= mod;
    if (typ) {
      odd_pairs++;
      odd_pairs %= mod;
      odd[1] += (odd_sum + mod - even_sum) % mod;
      odd[1] %= mod;
      odd[0] += (even_sum + mod - odd_sum) % mod;
      odd[0] %= mod;
    } else {
      ans += (2LL * ((odd_sum + mod - even_sum) % mod)) % mod;
      ans %= mod;
      ans +=
          ((((odd_sum + mod - even_sum) % mod) * even_pairs) % mod + even[1]) %
          mod;
      ans %= mod;
      ans +=
          ((((odd_sum + mod - even_sum) % mod) * even_pairs) % mod + even[1]) %
          mod;
      ans %= mod;
    }
  } else {
    even_sum += a[node];
    even_sum %= mod;
    if (typ) {
      even_pairs++;
      even_pairs %= mod;
      even[1] += (even_sum + mod - odd_sum) % mod;
      even[1] %= mod;
      even[0] += (odd_sum + mod - even_sum) % mod;
      even[0] %= mod;
    } else {
      ans += ((((even_sum + mod - odd_sum) % mod) * odd_pairs) % mod + odd[1]) %
             mod;
      ans %= mod;
      ans += ((((even_sum + mod - odd_sum) % mod) * odd_pairs) % mod + odd[1]) %
             mod;
      ans %= mod;
    }
  }
  for (int child : adj[node]) {
    if (child != par && !done[child]) {
      dfs(child, node, even_sum, odd_sum, h + 1, typ);
    }
  }
}
void solve(int centroid) {
  odd[0] = odd[1] = even[0] = even[1] = odd_pairs = even_pairs = 0;
  for (int child : adj[centroid]) {
    if (!done[child]) {
      dfs(child, centroid, 0, a[centroid], 2, 0);
      dfs(child, centroid, 0, 0, 1, 1);
    }
  }
}
void centroid_decomposition(int node) {
  calc_subsize(node, 0);
  int centroid = find_centroid(node, 0, sz[node] / 2);
  solve(centroid);
  done[centroid] = 1;
  for (int child : adj[centroid]) {
    if (!done[child]) {
      centroid_decomposition(child);
    }
  }
}
int main() {
  int n, u, v;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    a[i] += mod;
    a[i] %= mod;
    ans += a[i];
    ans %= mod;
  }
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  centroid_decomposition(1);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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 mod = 1e9 + 7;
const int N = 200005;
long long ans, odd_pairs, even_pairs, odd[2], even[2], done[N], sz[N], a[N];
vector<int> adj[N];
void calc_subsize(int node, int par) {
  sz[node] = 0;
  for (int child : adj[node]) {
    if (child != par && !done[child]) {
      calc_subsize(child, node);
      sz[node] += sz[child];
    }
  }
  sz[node]++;
}
int find_centroid(int node, int par, int conditional_size) {
  for (int child : adj[node]) {
    if (child != par && !done[child] && sz[child] > conditional_size) {
      return find_centroid(child, node, conditional_size);
    }
  }
  return node;
}
void dfs(int node, int par, int even_sum, int odd_sum, int h, int typ) {
  if (h & 1) {
    odd_sum += a[node];
    odd_sum %= mod;
    if (typ) {
      odd_pairs++;
      odd_pairs %= mod;
      odd[1] += (odd_sum + mod - even_sum) % mod;
      odd[1] %= mod;
      odd[0] += (even_sum + mod - odd_sum) % mod;
      odd[0] %= mod;
    } else {
      ans += (2LL * ((odd_sum + mod - even_sum) % mod)) % mod;
      ans %= mod;
      ans +=
          ((((odd_sum + mod - even_sum) % mod) * even_pairs) % mod + even[1]) %
          mod;
      ans %= mod;
      ans +=
          ((((odd_sum + mod - even_sum) % mod) * even_pairs) % mod + even[1]) %
          mod;
      ans %= mod;
    }
  } else {
    even_sum += a[node];
    even_sum %= mod;
    if (typ) {
      even_pairs++;
      even_pairs %= mod;
      even[1] += (even_sum + mod - odd_sum) % mod;
      even[1] %= mod;
      even[0] += (odd_sum + mod - even_sum) % mod;
      even[0] %= mod;
    } else {
      ans += ((((even_sum + mod - odd_sum) % mod) * odd_pairs) % mod + odd[1]) %
             mod;
      ans %= mod;
      ans += ((((even_sum + mod - odd_sum) % mod) * odd_pairs) % mod + odd[1]) %
             mod;
      ans %= mod;
    }
  }
  for (int child : adj[node]) {
    if (child != par && !done[child]) {
      dfs(child, node, even_sum, odd_sum, h + 1, typ);
    }
  }
}
void solve(int centroid) {
  odd[0] = odd[1] = even[0] = even[1] = odd_pairs = even_pairs = 0;
  for (int child : adj[centroid]) {
    if (!done[child]) {
      dfs(child, centroid, 0, a[centroid], 2, 0);
      dfs(child, centroid, 0, 0, 1, 1);
    }
  }
}
void centroid_decomposition(int node) {
  calc_subsize(node, 0);
  int centroid = find_centroid(node, 0, sz[node] / 2);
  solve(centroid);
  done[centroid] = 1;
  for (int child : adj[centroid]) {
    if (!done[child]) {
      centroid_decomposition(child);
    }
  }
}
int main() {
  int n, u, v;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    a[i] += mod;
    a[i] %= mod;
    ans += a[i];
    ans %= mod;
  }
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  centroid_decomposition(1);
  printf("%lld\n", ans);
  return 0;
}

```