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 double PI = atan(1.0) * 4;
const int64_t INF = 100000000000000003;
const int32_t LOG = 21;
const long long int BIG = pow(2, 20);
const long long int N = 100005;
long long int a[100005];
long long int cnt[100005];
long long int power(long long int x, long long int n) {
  if (n == 1) return x;
  if (n == 0) return 1;
  long long int temp = power(x, n / 2);
  temp = (temp * temp) % 1000000007;
  if (n & 1)
    return (temp * x) % 1000000007;
  else
    return temp;
}
long long int calc(long long int n, long long int x) {
  long long int i = 0;
  long long int pow = 0;
  while (cnt[i] % x == 0 && cnt[i] > 0) {
    cnt[i + 1] += cnt[i] / x;
    cnt[i] = 0;
    i++;
    pow++;
  }
  return min(pow, a[n - 1]);
}
int32_t main() {
  long long int n, x;
  cin >> n >> x;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int max1 = a[n - 1];
  for (long long int i = 0; i < n; i++) {
    if (max1 - a[i] < 100004) cnt[max1 - a[i]]++;
  }
  long long int added_power = calc(n, x);
  long long int sum = accumulate(a, a + n, 0LL) - a[n - 1];
  sum += added_power;
  cout << power(x, sum) << '\n';
  return 0;
}
