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 long long mod = 1000000007;
long long pw(long long base, long long pwr, long long x) {
  if (pwr == 0) return 1;
  if (pwr & 1LL)
    return (x * pw(((base * x) % mod), pwr - 1, x)) % mod;
  else {
    long long ret = pw(base, pwr >> 1LL, x);
    return (ret * ret) % mod;
  }
}
int main() {
  long long n, x;
  cin >> n >> x;
  long long arr[n], sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    sum += arr[i];
  }
  map<long long, long long> mp;
  for (int i = 0; i < n; i++) {
    long long tmp = sum - arr[i];
    mp[tmp]++;
  }
  map<long long, long long>::iterator it = mp.begin();
  while (it != mp.end()) {
    if (it->second >= x) {
      mp[it->first + 1] += it->second / x;
      it->second = it->second % x;
      it = mp.begin();
    } else
      it++;
  }
  long long ans = sum;
  for (auto it : mp)
    if (it.second != 0) ans = min(it.first, ans);
  cout << pw(x, ans, x);
  return 0;
}
