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 mod = 1e9 + 7;
long long pw(long long b, long long p) {
  if (!p) return 1;
  if (p == 1) return b;
  if (p % 2) return (b * pw((b * b) % mod, (p - 1) / 2) % mod) % mod;
  return pw((b * b) % mod, p / 2) % mod;
}
int main() {
  long long n, x;
  while (cin >> n >> x) {
    vector<long long> a(n);
    for (int i = 0; i < n; scanf("%I64d", &a[i++]))
      ;
    long long sum = 0;
    for (int i = 0; i < n; i++) sum += a[i];
    for (int i = 0; i < n; i++) a[i] = sum - a[i];
    sort(a.rbegin(), a.rend());
    long long res = -1;
    while (res == -1) {
      long long tmp = a.back(), cnt = 0;
      while (a.size() && a.back() == tmp) a.pop_back(), cnt++;
      if (cnt % x == 0)
        for (int i = 0; i < cnt / x; i++) a.push_back(tmp + 1);
      else
        res = pw(x, min(sum, tmp));
    }
    cout << res << endl;
  }
  return 0;
}
