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 INF = int(1e17);
const int sizeTree = 524289;
const double eps = 1e-7;
const double PI = 3.141592653589793238462;
const int N = 100010;
const int MOD = (int)1e9 + 7;
long long a[N];
int n;
long long x, sum, p;
map<long long, int> mp;
long long bin_pow(long long x, long long n) {
  long long ans = 1;
  long long tmp = x;
  while (n > 0) {
    if (n % 2 == 1) ans = ans * tmp % MOD;
    n /= 2;
    tmp = (tmp * tmp) % MOD;
  }
  return ans;
}
int main() {
  scanf("%d%I64d", &n, &x);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  sum = 0;
  for (int i = 1; i <= n; i++) sum += a[i];
  vector<long long> b;
  for (int i = 1; i <= n; i++) mp[sum - a[i]]++;
  map<long long, int>::iterator it = mp.begin();
  while (it != mp.end()) {
    if ((*it).second >= x) {
      long long k = (*it).second / x;
      (*it).second %= x;
      mp[(*it).first + 1] += k;
    }
    it++;
  }
  p = sum;
  for (map<long long, int>::const_iterator it = mp.begin(); it != mp.end();
       ++it) {
    if ((*it).second != 0) p = min(p, (*it).first);
  }
  printf("%I64d", bin_pow(x, p));
  return 0;
}
