You have unlimited number of coins with values 1, 2, …, n. You want to select some set of coins having the total value of S. 

It is allowed to have multiple coins with the same value in the set. What is the minimum number of coins required to get sum S?

Input

The only line of the input contains two integers n and S (1 ≤ n ≤ 100 000, 1 ≤ S ≤ 10^9)

Output

Print exactly one integer — the minimum number of coins required to obtain sum S.

Examples

Input

5 11


Output

3

Input

6 16


Output

3

Note

In the first example, some of the possible ways to get sum 11 with 3 coins are: 

  * (3, 4, 4)
  * (2, 4, 5)
  * (1, 5, 5)
  * (3, 3, 5) 



It is impossible to get sum 11 with less than 3 coins.

In the second example, some of the possible ways to get sum 16 with 3 coins are: 

  * (5, 5, 6)
  * (4, 6, 6) 



It is impossible to get sum 16 with less than 3 coins.
import java.io.*;
import java.util.*;
import java.math.*;

public class bhaa {

    InputStream is;
    PrintWriter o;
    /////////////////// CODED++ BY++ ++ ++ ++ BHAVYA++ ARORA++ ++ ++ ++ FROM++ JAYPEE++ INSTITUTE++ OF++ INFORMATION++ TECHNOLOGY++ ////////////////

///////////////////////// Make it work, make it right, make it fast. Make it work, make it right, make it fast. Make it work, make it right, make it fast. Make it work, make it right, make it fast. /////////////////

    boolean chpr(int n)
    {
    	if(n==1)
    	{
    		return true;
    	}if(n==2)
    	{
    		return true;
    	}
    	if(n==3)
    	{
    		return true;
    	}
    	if(n%2==0)
    	{
    		return false;

    	}
    	if(n%3==0)
    	{
    		return false;
    	}
    	
    	int w=2;
 		int i=5;
    	while(i*i<=n)
    	{
    		if(n%i==0)
    		{
    			return false;
    		}
    		i+=w;
    		w=6-w;
    	}
    	return true;

    }
    
    void solve() {


        int n=ni();
        int num=ni();
        int res=0;
        if(num%n!=0)
        {
            res=1;
        }
        res+=num/n;
        o.println(res);

    	

    }





   


    












    








































    //---------- I/O Template ----------
    
    public static void main(String[] args) { new bhaa().run(); }
    void run() { 
        is = System.in; 
        o = new PrintWriter(System.out);
        solve();
        o.flush();
    }
    
    byte input[] = new byte[1024];
    int len = 0, ptr = 0;
    
    int readByte() { 
        if(ptr >= len) { ptr = 0; 
            try { len = is.read(input); } 
            catch(IOException e) { throw new InputMismatchException(); } 
            if(len <= 0) { return -1; } 
        } return input[ptr++];
    }
    boolean isSpaceChar(int c) { return !( c >= 33 && c <= 126 ); }
    int skip() { 
        int b = readByte(); 
        while(b != -1 && isSpaceChar(b)) { b = readByte(); } 
        return b;
    }
    
    char nc() { return (char)skip(); }
    String ns() { 
        int b = skip(); 
        StringBuilder sb = new StringBuilder(); 
        while(!isSpaceChar(b)) { sb.appendCodePoint(b); b = readByte(); } 
        return sb.toString();
    }
    String nLine() { 
        int b = skip(); 
        StringBuilder sb = new StringBuilder(); 
        while( !(isSpaceChar(b) && b != ' ') ) { sb.appendCodePoint(b); b = readByte(); } 
        return sb.toString();
    }
    int ni() { 
        int n = 0, b = readByte(); 
        boolean minus = false; 
        while(b != -1 && !( (b >= '0' && b <= '9') || b == '-')) { b = readByte(); } 
        if(b == '-') { minus = true; b = readByte(); } 
        if(b == -1) { return -1; }  //no input 
        while(b >= '0' && b <= '9') { n = n * 10 + (b - '0'); b = readByte(); } 
        return minus ? -n : n;
    }
    long nl() { 
        long n = 0L;    int b = readByte(); 
        boolean minus = false; 
        while(b != -1 && !( (b >= '0' && b <= '9') || b == '-')) { b = readByte(); } 
        if(b == '-') { minus = true; b = readByte(); } 
        while(b >= '0' && b <= '9') { n = n * 10 + (b - '0'); b = readByte(); } 
        return minus ? -n : n;
    }

    double nd() { return Double.parseDouble(ns()); }
    float nf() { return Float.parseFloat(ns()); }
    int[] nia(int n) { 
        int a[] = new int[n]; 
        for(int i = 0; i < n; i++) { a[i] = ni(); } 
        return a;
    }
    long[] nla(int n) { 
        long a[] = new long[n]; 
        for(int i = 0; i < n; i++) { a[i] = nl(); } 
        return a;
    }
    int [][] nim(int n)
    {
        int mat[][]=new int[n][n];
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                mat[i][j]=ni();
            }
        }
        return mat;
    }
    long [][] nlm(int n)
    {
        long mat[][]=new long[n][n];
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                mat[i][j]=nl();
            }
        }
        return mat;
    }
    




    
    char[] ns(int n) { 
        char c[] = new char[n]; 
        int i, b = skip(); 
        for(i = 0; i < n; i++) { 
            if(isSpaceChar(b)) { break; } 
            c[i] = (char)b; b = readByte(); 
        } return i == n ? c : Arrays.copyOf(c,i);
    }
    void piarr(int arr[])
    {
        for(int i=0;i<arr.length;i++)
        {
            o.print(arr[i]+" ");

        }
        o.println();
    }
    void plarr(long arr[])
    {
        for(int i=0;i<arr.length;i++)
        {
            o.print(arr[i]+" ");

        }
        o.println();
    }
    
    void pimat(int mat[][])
    {
        for(int i=0;i<mat.length;i++)
        {
            for(int j=0;j<mat[0].length;j++)
            {
                o.print(mat[i][j]);
            }
            o.println();
        }
    }
    void plmat(long mat[][])
    {
        for(int i=0;i<mat.length;i++)
        {
            for(int j=0;j<mat[0].length;j++)
            {
                o.print(mat[i][j]);
            }
            o.println();
        }

    }




    //////////////////////////////////// template finished //////////////////////////////////////



}