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 M = 1e9 + 7;
long long a[100100], sum;
vector<long long> v;
long long pw(long long b, long long p) {
  if (!p) return 1;
  long long ret = pw(b, p / 2) % M;
  ret = (ret * ret) % M;
  if (p % 2) ret = ret * b % M;
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, first;
  cin >> n >> first;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = 0; i < n; i++) v.push_back(sum - a[i]);
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  while (true) {
    long long t = v.back();
    int cnt = 0;
    while (v.size() && t == v.back()) v.pop_back(), cnt++;
    if (cnt % first) return cout << pw(first, min(t, sum)) << endl, 0;
    for (int i = 0; i < cnt / first; i++) v.push_back(t + 1);
  }
}
