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 maxn = 2e5 + 10, mod = 1e9 + 7;
long long a[maxn];
long long bigmod(long long a, long long b, long long mod) {
  if (b == 0) return 1;
  long long x = bigmod(a, b / 2, mod);
  x = (x * x) % mod;
  if (b & 1) x = (x * a) % mod;
  return x;
}
int main() {
  long long n, x;
  cin >> n >> x;
  long long sum = 0;
  for (long long i = 1; i <= n; i++) cin >> a[i], sum += a[i];
  map<long long, long long> M;
  for (long long i = 1; i <= n; i++) {
    long long here = sum - a[i];
    M[here]++;
  }
  while (1) {
    auto it = *M.begin();
    if (it.second % x == 0) {
      M[it.first + 1] += it.second / x;
      M.erase(it.first);
    } else
      break;
  }
  auto p = *M.begin();
  cout << bigmod(x, min(p.first, sum), mod) << endl;
  return 0;
}
