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 mod = 1e9 + 7;
long long n, x, a[100005];
long long sm;
map<long long, long long> m;
long long pow_(long long p) {
  if (p == 0) return 1ll;
  if (p == 1) return x % mod;
  long long ret = pow_(p >> 1);
  ret = ret * ret % mod;
  if (p & 1) ret = ret * x % mod;
  return ret;
}
void solve() {
  scanf("%lld %lld", &n, &x);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1; i <= n; i++) sm = sm + a[i];
  for (long long i = 1; i <= n; i++) m[sm - a[i]]++;
  for (map<long long, long long>::iterator it = m.begin(); it != m.end();
       it++) {
    long long p = it->first, cnt = it->second;
    if (cnt % x != 0) {
      p = min(p, sm);
      printf("%lld", pow_(p));
      return;
    }
    m[++p] += cnt / x;
  }
  return;
}
signed main() {
  solve();
  return 0;
}
