Problem: Let's call a string adorable if its letters can be realigned in such a way that they form two consequent groups of equal symbols (note that different groups must contain different symbols). For example, ababa is adorable (you can transform it to aaabb, where the first three letters form a group of a-s and others — a group of b-s), but cccc is not since in each possible consequent partition letters in these two groups coincide.

You're given a string s. Check whether it can be split into two non-empty subsequences such that the strings formed by these subsequences are adorable. Here a subsequence is an arbitrary set of indexes of the string.

Input

The only line contains s (1 ≤ |s| ≤ 105) consisting of lowercase latin letters.

Output

Print «Yes» if the string can be split according to the criteria above or «No» otherwise.

Each letter can be printed in arbitrary case.

Examples

Input

ababa


Output

Yes


Input

zzcxx


Output

Yes


Input

yeee


Output

No

Note

In sample case two zzcxx can be split into subsequences zc and zxx each of which is adorable.

There's no suitable partition in sample case three.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
/*
 * 	Heart beats fast
 * 		Colors and promises
 * 			How to be brave
 * 				How can I love when I am afraid to fall...
 */
public class Main 
{
	static final int mod=1000000007;
	static final double eps=1e-8;
	static final long inf=100000000000000000L;
	static final boolean debug=true;
	static Reader in=new Reader();
	public static void main(String[] args) throws IOException
	{
		String a=ns();
		Set<Character> ho=new HashSet<Character>();
		int boolo[]=new int[26];
		for(int i=0; i<a.length(); i++)
		{
			ho.add(a.charAt(i));
			boolo[a.charAt(i)-'a']++;
		}
		if(ho.size()>4||a.length()<4)
			pr("No");
		else
		{
			if(ho.size()==4)
				pr("Yes");
			else if(ho.size()==3)
			{
				boolean pos=false;
				for(int i=0; i<26; i++)
					if(boolo[i]>1)
						pos=true;
				if(pos)
					pr("Yes");
				else
					pr("No");
			}
			else if(ho.size()==2)
			{
				int two=0;
				for(int i=0; i<26; i++)
					if(boolo[i]>1)
						two++;
				if(two>=2)
					pr("Yes");
				else
					pr("No");
			}
			else
				pr("No");
		}
		System.out.println(ans);
	}
	static StringBuilder ans=new StringBuilder();
	static long powm(long a, long b, long m)
	{
		long an=1;
		long c=a;
		while(b>0)
		{
			if(b%2==1)
				an=(an*c)%m;
			c=(c*c)%m;
			b>>=1;
		}
		return an;
	}
	static void pd(){if(debug)ans.append("hola");}
	static void pd(Object a){if(debug)ans.append(a+"\n");}
	static void pr(Object a){ans.append(a+"\n");}
	static void pr(){ans.append("\n");}
	static void p(Object a){ans.append(a);}
	static int ni(){return in.nextInt();}
	static long nl(){return in.nextLong();}
	static String ns(){return in.next();}
	static double nd(){return in.nextDouble();}
	static String pr(String a, long b)
	{
		String c="";
		while(b>0)
		{
			if(b%2==1)
				c=c.concat(a);
			a=a.concat(a);
			b>>=1;
		}
		return c;
	}
	static class Reader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public Reader() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
        	return Long.parseLong(next());
        }
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
    }
}