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.
from math import *
from copy import *
from string import *				# alpha = ascii_lowercase
from random import *
from sys import stdin
from sys import maxsize
from operator import *				# d = sorted(d.items(), key=itemgetter(1))
from itertools import *
from collections import Counter		# d = dict(Counter(l))
import math

def bin1(l,r,k,t,b,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		v=k**mid
		if(v==val):
			return v
		elif(v>val):
			ans=mid
			return bin1(mid+1,r,k,t,b,val,ans)
		else:
			return bin1(l,mid-1,k,t,b,val,ans)
		
def bin2(l,r,k,t,b,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		v=t*(k**mid)+b*(mid)
		if(v==val):
			return v
		elif(v>val):
			ans=mid
			return bin2(l,mid-1,k,t,b,val,ans)
		else:
			return bin2(mid+1,r,k,t,b,val,ans)

def SieveOfEratosthenes(n): 
      
    # Create a boolean array "prime[0..n]" and initialize 
    #  all entries it as true. A value in prime[i] will 
    # finally be false if i is Not a prime, else true. 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
          
        # If prime[p] is not changed, then it is a prime 
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    l=[]
    for i in range(2,n+1):
    	if(prime[i]):
    		l.append(i)
    return l
def bin(l,r,ll,val):
	if(l>r):
		return -1
	else:
		mid=(l+r)//2
		if(val>=ll[mid][0] and val<=ll[mid][1]):
			return mid
		elif(val<ll[mid][0]):
			return bin(l,mid-1,ll,val)
		else:
			return bin(mid+1,r,ll,val)
def deci(n):
	s=""
	while(n!=0):
		if(n%2==0):
			n=n//2
			s="0"+s
		else:
			n=n//2
			s="1"+s
	return s
def diff(s1,s2):
	if(len(s1)<len(s2)):
		v=len(s1)
		while(v!=len(s2)):
			s1="0"+s1
			v=v+1
	else:
		v=len(s2)
		while(v!=len(s1)):
			s2="0"+s2
			v=v+1
	c=0
	for i in range(len(s1)):
		if(s1[i:i+1]!=s2[i:i+1]):
			c=c+1
	return c
from sys import stdin, stdout 
def fac(a,b):
	v=a
	while(a!=b):
		v*=a-1
		a=a-1
	return v
def bino(l,r,n):
	if(l>r):
		return -1
	else:
		mid=(l+r)//2
		val1=math.log((n/mid)+1,2)
		val2=int(val1)
		if(val1==val2):
			return val1
		elif(val1<1.0):
			return bino(l,mid-1,n)
		else:
			return bino(mid+1,r,n)

def binary(l,r,ll,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		if(ll[mid]==val):
			return ll[mid]
		elif(ll[mid]<val):
			ans=ll[mid]
			return binary(mid+1,r,ll,val,ans)
		else:
			return binary(l,mid-1,ll,val,ans)
def check(n):
	v=1
	while(n!=0):
		if(n%10==0 or n%10==1):
			n=n//10
		else:
			v=0
			break
	return v

if __name__ == '__main__':
	n,s=map(int,input().split(" "))
	if(s>n):
		if(s%n==0):
			print(s//n)
		else:
			print(s//n+1)
	else:
		print(1)