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 expo(long long base, long long exponent, long long mod) {
  long long ans = 1;
  while (exponent != 0) {
    if ((exponent & 1) == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base = base * base;
    base %= mod;
    exponent >>= 1;
  }
  return ans % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long n, i, a[100005], s = 0, p, q, ans, x;
    cin >> n >> x;
    std::map<long long, long long> lol, hol;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      s += a[i];
    }
    for (i = 0; i < n; i++) a[i] = s - a[i];
    for (i = 0; i < n; i++) {
      lol[a[i]]++;
      hol[a[i]]++;
    }
    sort(a, a + n);
    p = a[0];
    q = lol[p];
    while (1) {
      if (q % x == 0) {
        lol[p + 1] += q / x;
        p = p + 1;
        q = lol[p];
      } else {
        ans = p;
        break;
      }
    }
    cout << expo(x, min(ans, s), 1000000007);
  }
  return 0;
}
