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 int maxn = 120000;
const int mod = 1000000007;
int n, x;
int a[maxn];
map<long long, int> mp;
long long pow(long long x, long long m) {
  long long ans = 1, now = x;
  while (m) {
    if (m & 1) ans = (ans * now) % mod;
    m >>= 1;
    now = (now * now) % mod;
  }
  return ans;
}
int main() {
  cin >> n >> x;
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    sum += a[i];
  }
  for (int i = 0; i < n; i++) {
    long long now = sum - a[i];
    map<long long, int>::iterator it = mp.find(now);
    if (it != mp.end()) {
      it->second++;
      while (it->second >= x) {
        it->second -= x;
        mp[++now]++;
        it = mp.find(now);
      }
    } else
      mp[now]++;
  }
  long long ans = 0;
  for (map<long long, int>::iterator it = mp.begin(); it != mp.end(); it++) {
    if (it->second != 0) {
      ans = it->first;
      break;
    }
  }
  ans = min(ans, sum);
  ans = pow(x, ans);
  cout << ans << endl;
  return 0;
}
