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).
import atexit
import io
import sys
# import os

# from bisect import *
# from collections import *
# from fractions import gcd
# from fractions import Fraction as fr_
# from itertools import *
# import math

inf = float('inf')  # sys.maxint in py2
inf_neg = float('-inf')  # sys.maxsize = 9*1e18
range_5 = int(1e5 + 1)
range_6 = int(1e6 + 1)
range_7 = int(1e7 + 1)
range_8 = int(1e8 + 1)
# sys.setrecursionlimit(range_8)

_INPUT_LINES = sys.stdin.read().splitlines()
input = iter(_INPUT_LINES).__next__
_OUTPUT_BUFFER = io.StringIO()
sys.stdout = _OUTPUT_BUFFER


@atexit.register
def write():
    sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())


def main():
    # n, k = map(int, input().split())
    n = int(input())
    c = 0
    while (n):
        n //= 2
        c += 1
    print(c)


if __name__ == '__main__':
    main()
