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 N = 1e5 + 2;
long long n, x, a[N];
map<long long, long long> m;
const long long mod = 1e9 + 7;
long long powmod(long long a, long long p) {
  if (p == 0) return 1;
  long long c = powmod(a, p / 2);
  if (p & 1) return (c * c) % mod * a % mod;
  return (c * c) % mod;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n >> x;
  long long sum = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (long long i = 0; i < n; i++) {
    m[sum - a[i]]++;
  }
  for (auto it = m.begin(); it != m.end(); it++) {
    long long k = (*it).first, l = (*it).second;
    if (l >= x) m[k + 1] += l / x;
    m[k] = l % x;
  }
  long long c;
  for (auto it : m) {
    if (it.second != 0) {
      c = it.first;
      break;
    }
  }
  sum = min(sum, c);
  cout << powmod(x, sum);
}
