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).
from sys import stdin, stdout 
from bisect import bisect_left, bisect_right
from collections import defaultdict
import math
from fractions import Fraction as frac
from random import random
cin = stdin.readline
def cout(x):
	stdout.write(str(x)+'\n')
def var(type = int):
    return type(stdin.readline())
def readline(type = int):
    return map(type,stdin.readline().split())
def readlist(type = int):
    return list(map(type,stdin.readline().split()))
def sorted_indexes(arr):
    return sorted(range(len(arr)),key=arr.__getitem__)
def printr(arr):
    [stdout.write(str(x)+' ')   for x in arr]
    cout('')
def find_lt(a, x):#'Find rightmost value less than x'
    i = bisect_left(a, x)
    if i:
        return a[i-1]
    raise ValueError
def find_gt(a, x):#'Find leftmost value greater than x'
    i = bisect_right(a, x)
    if i != len(a):
        return a[i]
    raise ValueError
def dist(x,y):
    return math.sqrt(x*x + y*y)
def binary_search(arr, x):
    i = bisect_left(arr, x)
    if i == len(arr) or arr[i] != x:
        return -1
    return i

# ---------------------Template ends-------------sdpt,sdpt131[Sudipta Banik]---------------------

# mp = [0]*201
# ops = [0]*201
# def go(arr,i,j ,dp):
#     if i==j and ops[i] is None:
#         return [mp[i],mp[i]]
#     if i>j:
#         return [0,0]
#     if dp[i][j]:
#         return dp[i][j]
#     mx = -1000000000
#     mn = 1000000000
#     for k in range(i+1,j,2):
#         if ops[k]:
#             left = go(arr,i,k-1,dp)
#             right = go(arr,k+1,j,dp)
#             mx = max(mx,left[0] + right[0])
#             mn = min(mn,left[1] + right[1])
#         else:
#             left = go(arr,i,k-1,dp)
#             right = go(arr,k+1,j,dp)
#             mx = max(mx,left[0] - right[1])
#             mn = min(mn,left[1] - right[0])
#     dp[i][j] = [mx,mn]
#     return [mx,mn]

    

n = var(int)

print(round(math.floor(math.log2(n)+1)))

    
    


    

    









# # def tobit(s):
# #     x = []
# #     for _ in range(3):
# #         x.append(s%2)
# #         s//=2
# #     return x

# def power(x, y, p) :
#     res = 1
#     x = x % p 
#     while (y > 0) :
#         if ((y & 1) == 1) :
#             res = (res * x) % p
#         y = y >> 1
#         x = (x * x) % p
#     return res




# def setCount(x):
#     return bin(x).count('1')

# for _ in range(var()):
    
#     a,b,n = readline()
#     p = 1000000007  
#     diff = abs(a-b)
#     mod  = power(a,n,p)
#     mod += power(b,n,p)
#     print(math.gcd(mod,diff)%p)
    
    
#     # n = var()
#     # # print(bin(n))
#     # if setCount(n)==2:
#     #     print(0)
#     # elif setCount(n)==1:
#     #     if n == 1:
#     #         print(2)
#     #     else:
#     #         print(1)
#     # elif setCount(n)==0:
#     #     print(3)
#     # else:
#     #     lo = n-1
#     #     hi = n+1
#     #     while(lo >= 3):
#     #         if(setCount(lo)==2):
#     #             break
#     #         lo -=1
#     #     while(hi <= 1000):
#     #         if(setCount(hi)==2):
#     #             break
#     #         hi +=1
        
#     #     if (hi - n) < (n - lo):
#     #         print(hi-n)
#     #     else:
#     #         print(n-lo)
        
    





# #     s = cin()
# #     t = cin()
# #     flg = False
# #     for el in range(8):
# #         bits = tobit(el)
# #         ch = []
# #         # printr(bits)
# #         for i in range(3):
# #             if bits[i]==0:
# #                 ch.append(s[i])
# #             else:
# #                 ch.append(t[i])
# #         ch.sort()
# #         if (''.join(ch) == 'bbo'):
# #             flg = True
# #             break
# #     if flg:
# #         print("yes")
# #     else:
# #         print("no")
