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.*;
import java.math.*;
import java.util.*;
public class ConstructivePrimeQuestion {  
 
	public static void main(String[] args) { 
		FastScanner I = new FastScanner(); //Input
		OutPut O = new OutPut(); //Output
		long N = I.nextLong();
		long bigPrime = N;
		ArrayList<Long> ans = new ArrayList<Long>();
		while (true) {
			boolean prime=true;
			for (int i = 2; i <= Math.sqrt(bigPrime); i++) {
				if (bigPrime%i==0) {
					prime=false;
					break;
				}
			}
			if (prime) {
				N-=bigPrime;
				ans.add(bigPrime);
				break;
			}
			bigPrime-=2; //Decrement wisely if still in loop
		}
		bigPrime=N;
		if (N>0) {
			while (true) {
				if (Prime(bigPrime)&&Prime(N-bigPrime)) {
					ans.add(bigPrime);
					if (N-bigPrime>0) ans.add(N-bigPrime);
					break;
				}
				bigPrime--; //Range is bound to be quite small now
			}
		}
		O.pln(ans.size());
		for (int i = 0; i < ans.size(); i++) O.p(ans.get(i)+" ");
	}
	public static boolean Prime(long N) {
		if (N==0) return true;
		if (N==1) return false;
		for (int i = 2; i <= Math.sqrt(N); i++) {
			if (N%i==0) return false;
		}
		return true;
	}
	public static long ceil(long num, long den) {long ans = num/den; if (num%den!=0) 
	ans++; return ans;}
	public static long GCD(long a, long b) {
		if (a==0||b==0) return Math.max(a,b);
		return GCD(Math.min(a, b),Math.max(a, b)%Math.min(a, b));
	}
	public static long FastExp(long base, long exp, long mod) {
		long ans=1;
		while (exp>0) {
			if (exp%2==1) ans*=base;
			exp/=2;
			base*=base;
			base%=mod;
			ans%=mod;
		}
		return ans;
	}
	public static long ModInv(long num,long mod) {return FastExp(num,mod-2,mod);}
	public static int pop(long x) { //Returns number of bits within a number
		int cnt = 0;
		while (x>0) {
			if (x%2==1) cnt++;
			x/=2;
		}
		return cnt;
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!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());};
	}
	static class OutPut{
		PrintWriter w = new PrintWriter(System.out);
		void pln(double x) {w.println(x);w.flush();}
		void pln(boolean x) {w.println(x);w.flush();}
		void pln(int x) {w.println(x);w.flush();}
		void pln(long x) {w.println(x);w.flush();}
		void pln(String x) {w.println(x);w.flush();}
		void pln(char x) {w.println(x);w.flush();}
		void pln(StringBuilder x) {w.println(x);w.flush();}
		void pln(BigInteger x) {w.println(x);w.flush();}
		void p(int x) {w.print(x);w.flush();}
		void p(long x) {w.print(x);w.flush();}
		void p(String x) {w.print(x);w.flush();}
		void p(char x) {w.print(x);w.flush();}
		void p(StringBuilder x) {w.print(x);w.flush();}
		void p(BigInteger x) {w.print(x);w.flush();}
		void p(double x) {w.print(x);w.flush();}
		void p(boolean x) {w.print(x);w.flush();}
	}
}