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 in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
const long long maxn = 1e6 + 10;
const long long mod = 1e9 + 7;
long long makh, a[maxn];
inline long long power(long long a, long long b) {
  return b == 0 ? 1 : power(a * a % mod, b >> 1) * (b % 2 ? a : 1) % mod;
}
int32_t main() {
  long long n = in(), p = in();
  for (long long i = 0; i < n; i++) {
    a[i] = in();
    makh += a[i];
  }
  vector<long long> vc;
  for (long long i = 0; i < n; i++) vc.push_back(makh - a[i]);
  sort(vc.begin(), vc.end(), greater<long long>());
  while (vc.size()) {
    long long number = vc.back();
    long long cnt = 0;
    while (vc.size() && vc.back() == number) {
      vc.pop_back();
      cnt++;
    }
    if (cnt % p == 0) {
      cnt /= p;
      while (cnt--) vc.push_back(number + 1);
    } else
      return cout << power(p, min(makh, number)) << endl, 0;
  }
}
