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.
import java.math.BigInteger;
import java.util.Scanner;

public class Main {
	public static void main (String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		long x = in.nextLong();
		BigInteger xB = BigInteger.valueOf(x);
		BigInteger p = BigInteger.valueOf(1000000007);
		long sum = 0;
		long[] a = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = in.nextLong();
			if (i + 1 < n) {
				sum += a[i];
			}
		}
		long curpow = 0;
		long curN = 0;
		boolean ok = true;
		for (int i = n-1; i >= 0; i--) {
			while (a[n-1] - a[i] != curpow) {
				if (curN % x == 0) {
					curpow++;
					curN /= x;
					if (curpow > a[n-1]) {
						ok = false;
						break;
					}
				} else {
					ok = false;
					break;
				}
			}
			if (!ok) {
				break;
			}
			curN++;
		}
		while (curN > 0 && curN % x == 0) {
			curpow++;
			curN /= x;
		}
		sum += Math.min(a[n-1], curpow);
		BigInteger pgcd = xB.modPow(BigInteger.valueOf(sum), p);
/*		if (eqN % x == 0) {
			BigInteger num = BigInteger.valueOf(0);
			BigInteger deno = xB.pow(a[n-1]);
			for (int i = 0; i < n; i++) {
				num = num.add(xB.modPow(BigInteger.valueOf(a[n-1]-a[i]),deno));
			}
			//System.out.println(num);
			//pgcd = pgcd.multiply(num).mod(p);
			pgcd = pgcd.multiply(num.gcd(deno).mod(p)).mod(p);
		}*/
		System.out.println(pgcd);
		in.close();
	}
}
