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.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Test {
    public static void main(String[] args) {

        FastReader in = new FastReader();
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        int i,j,a,b,c,n,k,l,m;
        String s,s1,s2;
        boolean is1 = false ,is = false,was = false;
        long sum = 0;
        int[] brr = new int[1];
        int[] arr = new int[1];
        //mergeSort(arr, 0, 7);
        //Arrays.binarySearch(arr,7);
        n = in.nextInt();

        a = n -2;

        if (a % 3 != 0){
            out.println(1 + " " + 1 + " " + a);
        }
        else {
            out.println(1 +  " " + 2 + " " + --a);
        }


        out.flush();


    }



    static void  mergeSort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Same as (l+r)/2, but avoids overflow for
            // large l and h
            int m = l+(r-l)/2;

            // Sort first and second halves
            mergeSort(arr, l, m);
            mergeSort(arr, m+1, r);

            merge(arr, l, m, r);
        }
    }


    static void merge(int arr[], int l, int m, int r)
    {
        int i, j, k;
        int n1 = m - l + 1;
        int n2 =  r - m;

        /* create temp arrays */
        int L[] = new int[n1],  R[] = new int[n2];

        /* Copy data to temp arrays L[] and R[] */
        for (i = 0; i < n1; i++)
            L[i] = arr[l + i];
        for (j = 0; j < n2; j++)
            R[j] = arr[m + 1+ j];

        /* Merge the temp arrays back into arr[l..r]*/
        i = 0; // Initial index of first subarray
        j = 0; // Initial index of second subarray
        k = l; // Initial index of merged subarray
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

    /* Copy the remaining elements of L[], if there
       are any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }

    /* Copy the remaining elements of R[], if there
       are any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }






    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}
