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 mod = 1000000000 + 7;
void fun() {}
long long bpow(long long x, long long v) {
  if (v == 0) return 1 % mod;
  if (v % 2 == 1) return (x * bpow(x, v - 1)) % mod;
  long long r = bpow(x, v / 2);
  return (r * r) % mod;
}
signed main() {
  fun();
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, x;
  cin >> n >> x;
  long long a[n];
  long long sum = 0;
  vector<long long> power;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (long long i = 0; i < n; i++) {
    power.push_back(sum - a[i]);
  }
  sort(power.rbegin(), power.rend());
  long long mn = power.back();
  long long i;
  while (1) {
    long long cnt = 0;
    mn = power.back();
    while (!power.empty() && power.back() == mn) {
      power.pop_back();
      cnt++;
    }
    if ((cnt % x) != 0) {
      if (mn > sum) mn = sum;
      cout << bpow(x, mn);
      return 0;
    }
    cnt = cnt / x;
    for (long long i = 0; i < cnt; i++) power.push_back(mn + 1);
  }
}
