Problem: One day Nikita found the string containing letters "a" and "b" only. 

Nikita thinks that string is beautiful if it can be cut into 3 strings (possibly empty) without changing the order of the letters, where the 1-st and the 3-rd one contain only letters "a" and the 2-nd contains only letters "b".

Nikita wants to make the string beautiful by removing some (possibly none) of its characters, but without changing their order. What is the maximum length of the string he can get?

Input

The first line contains a non-empty string of length not greater than 5 000 containing only lowercase English letters "a" and "b". 

Output

Print a single integer — the maximum possible size of beautiful string Nikita can get.

Examples

Input

abba


Output

4

Input

bab


Output

2

Note

It the first sample the string is already beautiful.

In the second sample he needs to delete one of "b" to make it beautiful.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class B {

	public static void main(String[] args)  {
		FastScanner scan = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		in = scan.next().toCharArray();
		n = in.length;
		dp = new Integer[n][3];
		out.println(go(0, 0));
		out.close();
	}
	
	static char[] in;
	static int n;
	static Integer[][] dp;
	
	static int go(int at, int i) {
		if(at == n) return 0;
		if(i == 3) return -0x3f3f3f3f;
		if(dp[at][i] != null) return dp[at][i];
		int res = 0;
		if(i == 0 || i == 2){
			if(in[at] == 'a'){
				res = Math.max(res, go(at+1, i)+1);
				res = Math.max(res, go(at+1, i+1)+1);
			}
		}
		if(i == 1) {
			if(in[at] == 'b'){
				res = Math.max(res, go(at+1, i)+1);
				res = Math.max(res, go(at+1, i+1)+1);
			}
		}
		res = Math.max(res, go(at+1, i));
		res = Math.max(res, go(at, i+1));
		return dp[at][i] = res;
	}
	
	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			try	{
				br = new BufferedReader(new InputStreamReader(System.in));
				st = new StringTokenizer(br.readLine());
			} catch (Exception e){e.printStackTrace();}
		}

		public String next() {
			if (st.hasMoreTokens())	return st.nextToken();
			try {st = new StringTokenizer(br.readLine());}
			catch (Exception e) {e.printStackTrace();}
			return st.nextToken();
		}

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

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

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

		public String nextLine() {
			String line = "";
			if(st.hasMoreTokens()) line = st.nextToken();
			else try {return br.readLine();}catch(IOException e){e.printStackTrace();}
			while(st.hasMoreTokens()) line += " "+st.nextToken();
			return line;
		}

		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for(int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}

		public long[] nextLongArray(int n){
			long[] a = new long[n];
			for(int i = 0; i < n; i++) a[i] = nextLong();
			return a;
		}

		public double[] nextDoubleArray(int n){
			double[] a = new double[n];
			for(int i = 0; i < n; i++) a[i] = nextDouble();
			return a;
		}

		public char[][] nextGrid(int n, int m){
			char[][] grid = new char[n][m];
			for(int i = 0; i < n; i++) grid[i] = next().toCharArray();
			return grid;
		}
	}

}