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 MAXN = (int)1e5 + 10;
const long long mod = 1000000007LL;
long long n, x, a[MAXN], b[MAXN];
long long power(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    y >>= 1;
    x = x * x % mod;
  }
  return res;
}
int main() {
  cin >> n >> x;
  long long tot = 0;
  for (int i = 0; i < n; i++) cin >> a[i], tot += a[i];
  for (int i = 0; i < n; i++) b[i] = tot - a[i];
  sort(b, b + n);
  long long ans = b[0];
  for (int i = 0; i < n; i++) b[i] -= ans;
  map<long long, int> te;
  for (int i = 0; i < n; i++) {
    te[b[i]]++;
    long long tmp = b[i];
    while (te[tmp] >= x) {
      te[tmp] -= x;
      te[tmp + 1]++;
      tmp++;
    }
  }
  long long add = 0;
  for (map<long long, int>::iterator it = te.begin(); it != te.end(); it++) {
    if ((*it).second != 0) {
      add = (*it).first;
      break;
    }
  }
  ans += add;
  ans = min(ans, tot);
  cout << power(x, ans) << endl;
  return 0;
}
