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.io.*;
import java.util.*;

public class C implements Runnable {

	final long MOD = 1000000000 + 7;
	
	private void solve() throws IOException {
		int n = nextInt();
		int[] a = new int[n];
		long x = nextLong(), sum = 0;
		HashMap<Long, Integer> map = new HashMap<Long, Integer>();
		PriorityQueue<Long> q = new PriorityQueue<Long>();
		for (int i = 0; i < n; i++) {
			a[i] = nextInt();
			sum += a[i];
		}
		for (int i = 0; i < n; i++) {
			long cur = sum - a[i];
			int k = 0;
			if (map.containsKey(cur)) k = map.remove(cur);
			map.put(cur, k + 1);
			if (k == 0) q.add(cur);
		}
		while (true) {
			long cur = q.poll();
			int cnt = map.get(cur), add = 0;
			while (cnt % x == 0) {
				cnt /= x;
				add++;
			}
			cur += add;
			int k = 0;
			if (map.containsKey(cur)) k = map.remove(cur);
			map.put(cur, k + cnt);
			if (k == 0) q.add(cur);
			if (add == 0) {
				q.add(cur);
				break;
			}
		}
		print(res(x, Math.min(q.peek(), sum)));
	}
	
	long res(long x, long pow) {
		if (pow == 0) return 1;
		if (pow == 1) return x;
		if (pow % 2 == 0) {
			long tmp = res(x, pow / 2);
			tmp %= MOD;
			return (tmp * tmp) % MOD;
		}
		long tmp = res(x, pow - 1);
		return (x * tmp) % MOD;
	}

	public static void main(String[] args) {
		new C().run();
	}

	BufferedReader reader;
	StringTokenizer tokenizer;
	PrintWriter writer;

	public void run() {
		try {
			reader = new BufferedReader(new InputStreamReader(System.in));
			writer = new PrintWriter(System.out);
			solve();
			writer.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(261);
		}
	}

	void halt() {
		writer.close();
		System.exit(0);
	}

	void print(Object... objects) {
		for (int i = 0; i < objects.length; i++) {
			if (i != 0)
				writer.print(' ');
			writer.print(objects[i]);
		}
	}

	void println(Object... objects) {
		print(objects);
		writer.println();
	}

	String nextLine() throws IOException {
		return reader.readLine();
	}

	String nextToken() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens())
			tokenizer = new StringTokenizer(nextLine());
		return tokenizer.nextToken();
	}

	int nextInt() throws NumberFormatException, IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws NumberFormatException, IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws NumberFormatException, IOException {
		return Double.parseDouble(nextToken());
	}
}
