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 java.util.*;
 
public class manathan_a extends Template{
    public static void main(String[] args) {
        int n; n = nI();
        int coins = 0; long sum = 0;
        while(n>0){
            n -= (sum+1);
            sum = 2*sum+1;
            coins ++;
        }
        pn(coins);
    }
}
 
class Template{
    static Scanner s = new Scanner(System.in);
    final static int nI(){return s.nextInt();}
    final static long nL(){return s.nextLong();}
    final static String nS(){return s.nextLine();}
    /* Renaming of some generic utility classes */
    final static class mylist extends ArrayList<pair>{}
    /* Constants Section */
    final static int INT_MIN = Integer.MIN_VALUE ;
    final static int INT_MAX = Integer.MAX_VALUE ;
    final static long LONG_MIN = Long.MIN_VALUE ;
    final static long LONG_MAX = Long.MAX_VALUE ;
    static long MOD = 1000000007 ;
    static void p(){System.out.println();}
    static void pn(long a){System.out.println(a);}
    static void pn(mylist a){System.out.println(a);}
    static void p(long a){System.out.printf("%d ", a);}
    static void p(long a, long b){System.out.printf("%d %d\n", a, b);}
    static void p(long a, long b, long c){System.out.printf("%d %d %d\n", a, b, c);}
    /* Implementation of operations modulo MOD */
    static long inv(long a){ return powM(a,MOD-2) ; }
    static long m(long a,long b){ return (a*b)%MOD ; }
    static long d(long a,long b){ return (a*inv(b))%MOD ; }
    static long powM(long a,long b){
        if(b<0) return powM(inv(a),-b) ;
        long ans=1 ;
        for( ; b!=0 ; a=(a*a)%MOD , b/=2) if((b&1)==1) ans = (ans*a)%MOD ;
        return ans ;
    }
}
class pair implements Comparable<pair>{
    long x ; long y ;
    pair(long _x,long _y){ x=_x ; y=_y ;}
    public int compareTo(pair p){
        return (this.x<p.x ? -1 : (this.x>p.x ? 1 : (this.y<p.y ? -1 : (this.y>p.y ? 1 : 0)))) ;
    }
    public String toString(){ return "(" + x + " " + y+ ")";}
}