Little C loves number «3» very much. He loves all things about it.

Now he has a positive integer n. He wants to split n into 3 positive integers a,b,c, such that a+b+c=n and none of the 3 integers is a multiple of 3. Help him to find a solution.

Input

A single line containing one integer n (3 ≤ n ≤ 10^9) — the integer Little C has.

Output

Print 3 positive integers a,b,c in a single line, such that a+b+c=n and none of them is a multiple of 3.

It can be proved that there is at least one solution. If there are multiple solutions, print any of them.

Examples

Input

3


Output

1 1 1

Input

233

Output

77 77 79
import java.util.*;
import java.io.*;

public class Main{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		long n=sc.nextLong();
		if(n==3){
			System.out.println(1+" "+1+" "+1);
		}else if(n==4){
			System.out.println(1+" "+2+" "+1);
		}else if(n==5){
			System.out.println(1+" "+2+" "+2);
		}else{
			long a=4;
			long b=1;
			long m=n-4;
			long c=m-b;
			// System.out.println(b+" "+c);
			while(b<n){
				if((b%3!=0) && (c%3!=0)){
					break;
				}else{
					b++;
					c=m-b;
				}
			}
			System.out.println(a+" "+b+" "+c);
		}
	}

	public static long nopair(ArrayList<Long> list1 , ArrayList<Long> list2){
		long count =0;
		for(int i=0;i<list1.size();i++){
			count =count + list1.get(i)*list2.get(i);
		}
		return count;
	}

	public static void ugly(int n){
		// if(n==1 || n==2 || n==3){
		// 	return null;
		// }else{
			int c=3;
			int i=3;
			while(c!=n){
				i++;
				if(i%2==0 || i%3==0 || i%5==0){
					c++;
				}
			}
		// }
	}

	public static ArrayList<Long> subarray(long[] array , int n){
		ArrayList<Long> list = new ArrayList<Long>();
		for(int i=1 ; i<n+1 ; i++){
			long value = array[i]*(i+1)*(n-i);
			list.add(value);
		}
		return list;
	}

	public static int count(int[] array , int n,int a, int b){
		int c=0;
		for(int i=a;i<=b;i++){
			if(n==array[i]){
				c++;
			}
		}
		return c;
	}
}