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 power(long long a, long long b) {
  if (b == 0) {
    return 1;
  } else if (b == 1) {
    return a;
  } else if (b % 2 == 0) {
    long long temp = power(a, b / 2);
    temp = temp * temp;
    temp = temp % 1000000007;
    return temp;
  } else {
    long long temp = power(a, ((b - 1) / 2));
    temp = temp * temp;
    temp = temp % 1000000007;
    temp = temp * a;
    temp = temp % 1000000007;
    return temp;
  }
}
int main() {
  long long n, x;
  cin >> n >> x;
  long long *a = new long long[n];
  long long maxi = LONG_LONG_MIN;
  long long sum = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    maxi = max(maxi, a[i]);
    sum = sum + a[i];
  }
  long long mini[n];
  map<long long, long long> ma;
  for (long long i = 0; i < n; i++) {
    mini[i] = sum - a[i];
    ma[mini[i]]++;
  }
  map<long long, long long>::iterator it;
  long long pp;
  for (it = ma.begin(); it != ma.end(); it++) {
    if ((*it).second % x == 0) {
      ma[(*it).first + 1] += (*it).second / x;
    } else {
      pp = (*it).first;
      break;
    }
  }
  pp = min(pp, sum);
  long long res = power(x, pp);
  cout << res << "\n";
}
