Simon has a prime number x and an array of non-negative integers a1, a2, ..., an.

Simon loves fractions very much. Today he wrote out number <image> on a piece of paper. After Simon led all fractions to a common denominator and summed them up, he got a fraction: <image>, where number t equals xa1 + a2 + ... + an. Now Simon wants to reduce the resulting fraction. 

Help him, find the greatest common divisor of numbers s and t. As GCD can be rather large, print it as a remainder after dividing it by number 1000000007 (109 + 7).

Input

The first line contains two positive integers n and x (1 ≤ n ≤ 105, 2 ≤ x ≤ 109) — the size of the array and the prime number.

The second line contains n space-separated integers a1, a2, ..., an (0 ≤ a1 ≤ a2 ≤ ... ≤ an ≤ 109). 

Output

Print a single number — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

2 2
2 2


Output

8


Input

3 3
1 2 3


Output

27


Input

2 2
29 29


Output

73741817


Input

4 5
0 0 0 0


Output

1

Note

In the first sample <image>. Thus, the answer to the problem is 8.

In the second sample, <image>. The answer to the problem is 27, as 351 = 13·27, 729 = 27·27.

In the third sample the answer to the problem is 1073741824 mod 1000000007 = 73741817.

In the fourth sample <image>. Thus, the answer to the problem is 1.
#include <bits/stdc++.h>
using namespace std;
long long first, sum, k, ans, a[100005];
vector<int> v;
map<long long, long long> m;
map<long long, long long>::iterator ii;
long long func(long long k) {
  if (k == 0) return 1;
  long long t = func(k / 2);
  t = t * t % 1000000007;
  if (k & 1) t = t * first % 1000000007;
  return t % 1000000007;
}
int main() {
  int n;
  cin >> n >> first;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    ans = ans + a[i];
  }
  for (int i = 0; i < n; ++i) m[ans - a[i]]++;
  for (ii = m.begin(); ii != m.end(); ++ii) {
    if (ii->second >= first) {
      k = ii->second / first;
      ii->second %= first;
      m[ii->first + 1] += k;
    }
    if (ii->second) {
      ans = min(ans, ii->first);
      break;
    }
  }
  cout << func(ans);
  return 0;
}
