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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<limits.h>

int max(int a, int b){
  return a >= b ? a : b;
}
int min(int a, int b){
  return b >= a ? a : b;
}
int sei(int a){
  return a>0 ? a : 0;
}
long int power(int a,int b){
  long int count=1;
  for(int i=0;i<b;i++){
    count *= 2;
  }
  return count;
}
int factorial(int n) {
    if (n > 0) return n*factorial(n - 1);
		else return 1;
}
int compare_up_int(const void *a, const void *b){
   return *(int*)a - *(int*)b;
}
int compare_down_int(const void *a, const void *b){
   return *(int*)b - *(int*)a;
}
int euclid(int a, int b){
  if(a < b){int tmp = a;a = b;b = tmp;}
  int r = a % b;
  if(r < 0) r += b;
  while(r != 0){
    a = b;b = r;r = a % b;
    if(r < 0) r += b;
  }
  return b;
}

int main(){
  long int a,b,count=0;
  scanf("%ld %ld",&a,&b);
  for(long int i=1;i<b;i++){
    if(a*(long int)power(2,i) > b){
      count = i;break;
    }    
    if(a*(long int)power(2,i) >= b){
      count=i+1;break;
    }
  }
  printf("%d",count);
  return 0;
}