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, long long mod) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = ((ans % mod) * (a % mod)) % mod;
    ;
    a = ((a % mod) * (a % mod)) % mod;
    b = b / 2;
  }
  return ans;
}
void func() {
  long long n, x;
  cin >> n >> x;
  long long a[n], sum = 0;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    sum += a[i];
  }
  vector<long long> v;
  for (long long i = 0; i < n; ++i) {
    v.push_back(sum - a[i]);
  }
  sort(v.rbegin(), v.rend());
  while (true) {
    long long m = *v.rbegin();
    long long cnt = 0;
    while (v.size() > 0 && (*v.rbegin() == m)) {
      ++cnt;
      v.pop_back();
    }
    if (cnt % x != 0) {
      m = min(m, sum);
      long long ans = power(x, m, 1000000007);
      cout << ans << "\n";
      return;
    } else {
      cnt = cnt / x;
      for (long long i = 0; i < cnt; ++i) {
        v.push_back(m + 1);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    func();
  }
}
