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.*;


public class test {
    int INF = (int)1e9;
    long MOD = 1000000007;
    
    
    
    void solve(InputReader in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        if((n-2)%3==0) {
            out.println("1 2 "+(n-3));
        } else {
            out.println("1 1 "+(n-2));
        }
    }
    
    
    public static void main(String[] args) throws IOException {
        if(args.length>0 && args[0].equalsIgnoreCase("d")) {
            DEBUG_FLAG = true;
        }
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = 1;//in.nextInt();
        long start = System.nanoTime();
        while(t-->0) {
            new test().solve(in, out);
        }
        long end = System.nanoTime();
        debug("\nTime: " + (end-start)/1e6 + " \n\n");
        out.close();
    }
    
    static boolean DEBUG_FLAG = false;
    static void debug(String s) {
        if(DEBUG_FLAG) System.out.print(s);
    }
    
    public static void show(Object... o) {
        System.out.println(Arrays.deepToString(o));
    }
    
    static class InputReader {
        static BufferedReader br;
        static StringTokenizer st;
    
        public InputReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                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());
        }
    }
}