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;
const int mod = 1000000007;
long long num[100000 + 50];
long long pow_mod(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) {
      ret = ret * a % mod;
    }
    b = b >> 1;
    a = a * a % mod;
  }
  return ret;
}
int main() {
  long long n, x;
  map<long long, long long> m;
  map<long long, long long>::iterator it;
  cin >> n >> x;
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    cin >> num[i];
    sum += num[i];
  }
  for (int i = 1; i <= n; i++) {
    m[num[n] - num[i]]++;
  }
  while (1) {
    it = m.begin();
    long long a = it->first;
    long long b = it->second;
    if (b % x == 0) {
      while (b % x == 0) {
        b = b / x;
        a++;
      }
      m.erase(it);
      m[a] += b;
    } else {
      break;
    }
  }
  cout << pow_mod(x, sum - num[n] + min(num[n], m.begin()->first));
  return 0;
}
