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;
map<long long, long long> ais;
long long fastexpo(long long base, long long pw) {
  if (pw == 0) return 1ll;
  if (pw == 1) return base;
  long long res = fastexpo(base, pw / 2);
  res *= res;
  res %= 1000000007;
  if (pw % 2 == 1) res = (res * base) % 1000000007;
  return res;
}
int main() {
  long long n, x, ai;
  cin >> n >> x;
  long long tot = 0;
  map<long long, long long>::iterator it;
  for (int i = 0; i < n; ++i) {
    cin >> ai;
    tot += ai;
    ai *= -1ll;
    it = ais.find(ai);
    if (it == ais.end()) {
      ais[ai] = 1;
    } else {
      ++ais[ai];
    }
  }
  it = ais.begin();
  long long nitf, tmp;
  while (it != ais.end()) {
    nitf = -(it->first);
    if ((it->second) % x > 0 || (nitf < 0)) {
      cout << fastexpo(x, min(tot - nitf, tot));
      return 0;
    }
    tmp = (it->second) / x;
    (ais.find(-nitf + 1) != ais.end()) ? ais[-nitf + 1] += tmp
                                       : ais[-nitf + 1] = tmp;
    ++it;
  }
  return 0;
}
