Problem: Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at most than three primes.

More formally, you are given an odd numer n. Find a set of numbers pi (1 ≤ i ≤ k), such that

  1. 1 ≤ k ≤ 3
  2. pi is a prime
  3. <image>



The numbers pi do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.

Input

The single line contains an odd number n (3 ≤ n < 109).

Output

In the first line print k (1 ≤ k ≤ 3), showing how many numbers are in the representation you found.

In the second line print numbers pi in any order. If there are multiple possible solutions, you can print any of them.

Examples

Input

27


Output

3
5 11 11

Note

A prime is an integer strictly larger than one that is divisible only by one and by itself.

Solution (in JAVA):

Solution:
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;

public class R324qD {

	public static void main(String args[]) {
		
		InputReader in = new InputReader(System.in);
		int n = in.nextInt();
		
		if(isPrime(n)){
			System.out.println(1);
			System.out.println(n);
			return;
		}
		
		if(isPrime(n-2)){
			System.out.println(2);
			System.out.println(2 + " " + (n-2));
			return;
		}
		
		boolean prime[] = sieve(2000001);
		
		for(int i=2;i<=500;i++){
			for(int j=i;j<=500;j++){
				int k = n - i - j;
				if(k >= 1 && prime[i] && prime[j] && isPrime(k) && (i+j+k) == n){
					System.out.println(3);
					System.out.println(i + " " + j + " " + k);
					return;
				}
			}
		}
		
		for(int i=3;i<=2000000;i+=2){
			if(prime[i] && (n-i-3) >=1 && isPrime(n-i-3)){
				System.out.println(3);
				System.out.println(3 + " " + i + " " + (n-i-3));
				return;
			}
		}
		
		throw new RuntimeException();
	}
	
	public static boolean[] sieve(int n){
		boolean prime[] = new boolean[n+1];
		Arrays.fill(prime,true);
		prime[0] = prime[1] = false;
		
		int m = (int) Math.sqrt(n) + 1;
		for(int i=2;i<=m;i++)
			if(prime[i]) 
				for(int k=i*i;k<=n;k+=i) 
					prime[k] = false;
		
		return prime;
	}
	
	public static boolean isPrime(int n){
		if(n <= 1) return false;
		if(n == 2) return true;
		if(n % 2 == 0) return false;
		int m = (int) (Math.sqrt(n) + 15); //2 is not a factor coz we would have returned false then
		m = Math.min(m, n-1);
		for(int i = 3;i <= m;i += 2) //if a*b = n then atleat one of a,b <= m
			if(n % i == 0) return false;
		return true;
	}

	static class InputReader {

		private InputStream stream;
		private byte[] buf = new byte[8192];
		private int curChar;
		private int snumChars;
		private SpaceCharFilter filter;

		public InputReader(InputStream stream) {
			this.stream = stream;
		}

		public int snext() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public int nextInt() {
			int c = snext();
			while (isSpaceChar(c))
				c = snext();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = snext();
			}

			int res = 0;

			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = snext();
			} while (!isSpaceChar(c));

			return res * sgn;
		}
		
		public long nextLong() {
			int c = snext();
			while (isSpaceChar(c))
				c = snext();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = snext();
			}

			long res = 0;

			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = snext();
			} while (!isSpaceChar(c));

			return res * sgn;
		}
		
		public String readString() {
			int c = snext();
			while (isSpaceChar(c))
				c = snext();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = snext();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public boolean isSpaceChar(int c) {
			if (filter != null)
				return filter.isSpaceChar(c);
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
	}

}