The Little Elephant has an integer a, written in the binary notation. He wants to write this number on a piece of paper.

To make sure that the number a fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number a in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).

The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.

Input

The single line contains integer a, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.

Output

In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.

Examples

Input

101


Output

11


Input

110010


Output

11010

Note

In the first sample the best strategy is to delete the second digit. That results in number 112 = 310.

In the second sample the best strategy is to delete the third or fourth digits — that results in number 110102 = 2610.
/*input
110010


*/
import java.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
	static PrintWriter out;
	static int MOD = 1000000007;
	static FastReader scan;

    /*-------- I/O using short named function ---------*/
	public static String ns(){return scan.next();}
	public static int ni(){return scan.nextInt();}
	public static long nl(){return scan.nextLong();}
	public static double nd(){return scan.nextDouble();}
	public static String nln(){return scan.nextLine();}
	public static void p(Object o){out.print(o);}
	public static void ps(Object o){out.print(o + " ");}
	public static void pn(Object o){out.println(o);}
    /*-------- for output of an array ---------------------*/
     static void iPA(int arr []){
     	 StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
	static void lPA(long arr []){
	     StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
	static void sPA(String arr []){
	     StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
	static void dPA(double arr []){
	     StringBuilder output = new StringBuilder();
	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
	}
    /*-------------- for input in an array ---------------------*/
	static void iIA(int arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = ni();
	}
	static void lIA(long arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = nl(); 
	}
	static void sIA(String arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = ns(); 
	}
	static void dIA(double arr[]){
	     for(int i=0; i<arr.length; i++)arr[i] = nd();
	}
	/*------------ for taking input faster ----------------*/
     static class FastReader
	{
		BufferedReader br;
		StringTokenizer st;

		public FastReader()
		{
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next()
		{
			while (st == null || !st.hasMoreElements())
			{
				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());
		}

		String nextLine()
		{
			String str = "";
			try
			{
				str = br.readLine();
			}
			catch (IOException e)
			{
				e.printStackTrace();
			}
			return str;
		}
	}

	// Method to check if x is power of 2
    static boolean isPowerOfTwo (int x) { return x!=0 && ((x&(x-1)) == 0);}
    //Method to return lcm of two numbers
    static int gcd(int a, int b){return a==0?b:gcd(b % a, a); }
    // method to return LCM of two numbers
    static int lcm(int a, int b){return (a / gcd(a, b)) * b;}
    //Method to count digit of a number
    static int countDigit(long n){return (int)Math.floor(Math.log10(n) + 1);}

    //Method to find the max in an array
    static int getMax(int arr[]){
    	int max = arr[0];
    	for(int i=0; i<arr.length; i++){
    		max = arr[i]>max?arr[i]:max;
    	}
    	return max;
    }
    //Method to find the min in an array
    static int getMin(int arr[]){
    	int min = arr[0];
    	for(int i=0; i<arr.length; i++){
    		min = min>arr[i]?arr[i]:min;
    	}
    	return min;
    }
    
	

	public static void main (String[] args) throws java.lang.Exception
	{
		OutputStream outputStream =System.out;
        out =new PrintWriter(outputStream);
        scan =new FastReader();
		//for fast output sometimes
		StringBuilder sb = new StringBuilder();
		String s = ns();
		int n = s.length();
		boolean found = false;
		for(int i=0; i<s.length()-1; i++){
			char c = s.charAt(i);
			if(found){
				sb.append(c);
			}
			else{
				if(c == '1')
					sb.append(c);
				else{
					found = true;
				}
			}
			if(c == '0'){
				found = true;
			}
			

		}
		if(found){
			sb.append(s.charAt(n-1));
		}
		pn(sb);
		out.close();
	}
	
}
