#include <bits/stdc++.h>
using namespace std;
long long modulo(long long base, long long exponent, long long modulus);
long long choose(long long n, long long k);
long long inverse(long long a, long long m);
void build();
long long ncr(long long n, long long r);
vector<long long> fact(1000010);
long long odd[1000010];
long long even[1000010];
long long sum(long long a, long long b) { return (a + b) % 1000000007; }
long long sub(long long a, long long b) {
  return (a - b + 1000000007) % 1000000007;
}
long long mult(long long a, long long b) { return (a * b) % 1000000007; }
long long ans;
vector<long long> a;
vector<long long> adj[1000010];
long long dist[1000010];
void dfs(long long x, long long p) {
  long long c = 0;
  for (int i = 0; i < adj[x].size(); i++) {
    long long y = adj[x][i];
    if (p == y) continue;
    dist[y] = dist[x] + 1;
    dfs(y, x);
    c++;
    if (c > 0) {
      ans = sum(ans, mult(mult(mult(odd[x], even[y]), a[x]), 2));
      ans = sum(ans, mult(mult(mult(even[x], odd[y]), -a[x]), 2));
    }
    odd[x] = sum(odd[x], even[y]);
    even[x] = sum(even[x], odd[y]);
  }
  odd[x] += 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  a.resize(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < (n - 1); i++) {
    long long x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    long long x = i;
    long long tod, teve;
    if (dist[i] % 2) {
      tod = even[1];
      teve = odd[1];
      ans =
          ((ans + 2 *
                      (odd[i] * (even[1] - odd[i] + 1) % 1000000007 -
                       even[i] * (odd[1] - even[i]) % 1000000007 + 1000000007) %
                      1000000007 * a[i] % 1000000007) %
               1000000007 +
           1000000007) %
          1000000007;
    } else {
      tod = odd[1];
      teve = even[1];
      ans = ((ans +
              2 *
                  (odd[i] * (odd[1] - odd[i] + 1) % 1000000007 -
                   even[i] * (even[1] - even[i]) % 1000000007 + 1000000007) %
                  1000000007 * a[i] % 1000000007) %
                 1000000007 +
             1000000007) %
            1000000007;
    }
    ans = sub(ans, a[x]);
  }
  cout << ans << endl;
  return 0;
}
long long ncr(long long n, long long r) {
  if (r > n || r < 0 || n < 0) return 0;
  long long ans = fact[n];
  long long temp = (fact[n - r] * fact[r]) % 1000000007;
  ans = (ans * inverse(temp, 1000000007)) % 1000000007;
  return ans;
}
void build() {
  fact[0] = 1;
  for (long long i = 1; i < 1000010; i++)
    fact[i] = (fact[i - 1] * i) % 1000000007;
}
long long modulo(long long base, long long exponent, long long modulus) {
  if (modulus == 1) return 0;
  long long result = 1;
  base = base % modulus;
  while (exponent > 0) {
    if (exponent % 2 == 1) {
      result = (result * base) % modulus;
    }
    exponent = exponent >> 1;
    base = (base * base) % modulus;
  }
  return result;
}
long long choose(long long n, long long k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
void EE(long long a, long long b, long long &co1, long long &co2) {
  if (a % b == 0) {
    co1 = 0;
    co2 = 1;
    return;
  }
  EE(b, a % b, co1, co2);
  long long temp = co1;
  co1 = co2;
  co2 = temp - co2 * (a / b);
}
long long inverse(long long a, long long m) {
  long long x, y;
  EE(a, m, x, y);
  if (x < 0) x += m;
  return x;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long modulo(long long base, long long exponent, long long modulus);
long long choose(long long n, long long k);
long long inverse(long long a, long long m);
void build();
long long ncr(long long n, long long r);
vector<long long> fact(1000010);
long long odd[1000010];
long long even[1000010];
long long sum(long long a, long long b) { return (a + b) % 1000000007; }
long long sub(long long a, long long b) {
  return (a - b + 1000000007) % 1000000007;
}
long long mult(long long a, long long b) { return (a * b) % 1000000007; }
long long ans;
vector<long long> a;
vector<long long> adj[1000010];
long long dist[1000010];
void dfs(long long x, long long p) {
  long long c = 0;
  for (int i = 0; i < adj[x].size(); i++) {
    long long y = adj[x][i];
    if (p == y) continue;
    dist[y] = dist[x] + 1;
    dfs(y, x);
    c++;
    if (c > 0) {
      ans = sum(ans, mult(mult(mult(odd[x], even[y]), a[x]), 2));
      ans = sum(ans, mult(mult(mult(even[x], odd[y]), -a[x]), 2));
    }
    odd[x] = sum(odd[x], even[y]);
    even[x] = sum(even[x], odd[y]);
  }
  odd[x] += 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  a.resize(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < (n - 1); i++) {
    long long x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    long long x = i;
    long long tod, teve;
    if (dist[i] % 2) {
      tod = even[1];
      teve = odd[1];
      ans =
          ((ans + 2 *
                      (odd[i] * (even[1] - odd[i] + 1) % 1000000007 -
                       even[i] * (odd[1] - even[i]) % 1000000007 + 1000000007) %
                      1000000007 * a[i] % 1000000007) %
               1000000007 +
           1000000007) %
          1000000007;
    } else {
      tod = odd[1];
      teve = even[1];
      ans = ((ans +
              2 *
                  (odd[i] * (odd[1] - odd[i] + 1) % 1000000007 -
                   even[i] * (even[1] - even[i]) % 1000000007 + 1000000007) %
                  1000000007 * a[i] % 1000000007) %
                 1000000007 +
             1000000007) %
            1000000007;
    }
    ans = sub(ans, a[x]);
  }
  cout << ans << endl;
  return 0;
}
long long ncr(long long n, long long r) {
  if (r > n || r < 0 || n < 0) return 0;
  long long ans = fact[n];
  long long temp = (fact[n - r] * fact[r]) % 1000000007;
  ans = (ans * inverse(temp, 1000000007)) % 1000000007;
  return ans;
}
void build() {
  fact[0] = 1;
  for (long long i = 1; i < 1000010; i++)
    fact[i] = (fact[i - 1] * i) % 1000000007;
}
long long modulo(long long base, long long exponent, long long modulus) {
  if (modulus == 1) return 0;
  long long result = 1;
  base = base % modulus;
  while (exponent > 0) {
    if (exponent % 2 == 1) {
      result = (result * base) % modulus;
    }
    exponent = exponent >> 1;
    base = (base * base) % modulus;
  }
  return result;
}
long long choose(long long n, long long k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
void EE(long long a, long long b, long long &co1, long long &co2) {
  if (a % b == 0) {
    co1 = 0;
    co2 = 1;
    return;
  }
  EE(b, a % b, co1, co2);
  long long temp = co1;
  co1 = co2;
  co2 = temp - co2 * (a / b);
}
long long inverse(long long a, long long m) {
  long long x, y;
  EE(a, m, x, y);
  if (x < 0) x += m;
  return x;
}

```