Problem:
Score : 300 points

Problem Statement

As a token of his gratitude, Takahashi has decided to give his mother an integer sequence.
The sequence A needs to satisfy the conditions below:

A consists of integers between X and Y (inclusive).

For each 1\leq i \leq |A|-1, A_{i+1} is a multiple of A_i and strictly greater than A_i.

Find the maximum possible length of the sequence.

Constraints

1 \leq X \leq Y \leq 10^{18}

All input values are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

Print the maximum possible length of the sequence.

Sample Input 1

3 20

Sample Output 1

3

The sequence 3,6,18 satisfies the conditions.

Sample Input 2

25 100

Sample Output 2

3

Sample Input 3

314159265 358979323846264338

Sample Output 3

31

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <limits.h>	
#define inf (INT_MAX-1)
#define INF 9223372036854775807
#define sq(n) ((n)*(n))
#define rep(i,n) for(i=0;i<n;i++)
#define rev(i,n) for(i=n-1;i>=0;i--)
#define sort(a,n) qsort(a,n,sizeof(TYPE),cmp)
#define sort_r(a,n) qsort(a,n,sizeof(TYPE),cmp_r);
#define TYPE int
#define MEMSET(a) memset(a,0,sizeof(a))
const int mod=(int)1e09+7;

int in(void){
	int i;scanf("%d",&i);
	return i;
}
long long llin(void){
	long long i;scanf("%lld",&i);
	return i;
}
double din(void){
	double i;scanf("%lf",&i);
	return i;
}
void chin(char s[]){
	scanf("%s",s);
}
void print(int a){
	printf("%d\n",a);
}
void llprint(long long a){
	printf("%lld\n",a);
}
void dprint(double a){
	printf("%.10f\n",a);
}
void print2(int a,int b){
	printf("%d %d\n",a,b);
}
long long max(long long a,long long b){
	return a>b?a:b;
}
long long min(long long a,long long b){
	return a<b?a:b;
}
long long llabs(long long a){
	return a>0?a:-a;
}
double dmax(double a,double b){
	return a>b?a:b;
}
int cmp(const void *a,const void *b){
	return *(TYPE *)a-*(TYPE *)b;
}
int cmp_r(const void *a,const void *b){
	return *(TYPE *)b-*(TYPE *)a;
}
void swap(int *a,int *b){
	int t=*a;
	*a=*b;
	*b=t;
}

int main(void){
	long long x=llin(),y=llin(),ans=0,n;
	n=x;
	while(n<=y){
		ans++;
		n*=2;
	}
	llprint(ans);
	return 0;
}