To become the king of Codeforces, Kuroni has to solve the following problem.

He is given n numbers a_1, a_2, ..., a_n. Help Kuroni to calculate ∏_{1≤ i<j≤ n} |a_i - a_j|. As result can be very big, output it modulo m.

If you are not familiar with short notation, ∏_{1≤ i<j≤ n} |a_i - a_j| is equal to |a_1 - a_2|⋅|a_1 - a_3|⋅ ... ⋅|a_1 - a_n|⋅|a_2 - a_3|⋅|a_2 - a_4|⋅ ... ⋅|a_2 - a_n| ⋅ ... ⋅ |a_{n-1} - a_n|. In other words, this is the product of |a_i - a_j| for all 1≤ i < j ≤ n.

Input

The first line contains two integers n, m (2≤ n ≤ 2⋅ 10^5, 1≤ m ≤ 1000) — number of numbers and modulo.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).

Output

Output the single number — ∏_{1≤ i<j≤ n} |a_i - a_j| mod m.

Examples

Input


2 10
8 5


Output


3

Input


3 12
1 4 5


Output


0

Input


3 7
1 4 9


Output


1

Note

In the first sample, |8 - 5| = 3 ≡ 3 mod 10.

In the second sample, |1 - 4|⋅|1 - 5|⋅|4 - 5| = 3⋅ 4 ⋅ 1 = 12 ≡ 0 mod 12.

In the third sample, |1 - 4|⋅|1 - 9|⋅|4 - 9| = 3 ⋅ 8 ⋅ 5 = 120 ≡ 1 mod 7.
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
	    Scanner sc=new Scanner(System.in);
	   // BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    int t=1;
	   // t=sc.nextInt();
	    //int t=Integer.parseInt(br.readLine());
	    while(--t>=0){
	        int n=sc.nextInt();
	        long a[]=new long[n];
	        long m=sc.nextLong();
	        for(int i=0;i<n;i++)a[i]=sc.nextLong();
	        if(n>(m+1)){
	            System.out.println(0);
	            continue;
	        }
	        long ans=1;
	        for(int i=0;i<n;i++){
	            for(int j=i+1;j<n;j++){
	                ans=(ans%m*Math.abs(a[i]-a[j])%m)%m;
	                
	            }
	            
	        }
	     System.out.println(ans);
	        
	    }
	    
	}
}