You have n coins, each of the same value of 1.

Distribute them into packets such that any amount x (1 ≤ x ≤ n) can be formed using some (possibly one or all) number of these packets.

Each packet may only be used entirely or not used at all. No packet may be used more than once in the formation of the single x, however it may be reused for the formation of other x's.

Find the minimum number of packets in such a distribution.

Input

The only line contains a single integer n (1 ≤ n ≤ 10^9) — the number of coins you have.

Output

Output a single integer — the minimum possible number of packets, satisfying the condition above.

Examples

Input

6


Output

3

Input

2


Output

2

Note

In the first example, three packets with 1, 2 and 3 coins can be made to get any amount x (1≤ x≤ 6).

  * To get 1 use the packet with 1 coin. 
  * To get 2 use the packet with 2 coins. 
  * To get 3 use the packet with 3 coins. 
  * To get 4 use packets with 1 and 3 coins. 
  * To get 5 use packets with 2 and 3 coins 
  * To get 6 use all packets. 



In the second example, two packets with 1 and 1 coins can be made to get any amount x (1≤ x≤ 2).
#_________________ Mukul Mohan Varshney _______________#
 
#Template
import sys
import os
import math
import copy
from math import gcd
from bisect import bisect
from io import BytesIO, IOBase
from math import sqrt,floor,factorial,gcd,log,ceil
from collections import deque,Counter,defaultdict
from itertools import permutations, combinations
from itertools import accumulate
 
#define function 
def Int(): return int(sys.stdin.readline())
def Mint(): return map(int,sys.stdin.readline().split())
def Lstr(): return list(sys.stdin.readline().strip())
def Str(): return sys.stdin.readline().strip()
def Mstr(): return map(str,sys.stdin.readline().strip().split())
def List(): return list(map(int,sys.stdin.readline().split()))
def Hash(): return dict()
def Mod(): return 1000000007
def Ncr(n,r,p): return ((fact[n])*((ifact[r]*ifact[n-r])%p))%p
def Most_frequent(list): return max(set(list), key = list.count)
def Mat2x2(n): return [List() for _ in range(n)]
def Lcm(x,y):  return (x*y)//gcd(x,y)
def dtob(n): return bin(n).replace("0b","")
def btod(n): return int(n,2) 
def common(l1, l2): 
    return set(l1).intersection(l2) 
         

# Driver Code     
def solution():
     #for _ in range(Int()):
          n=Int()
          c=0
          k=1
          while(n>0):
               n=n-k
               c+=1
               k=k*2
          print(c)     


#Call the solve function          
if __name__ == "__main__":
     solution()  