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>
 /*
int Sumed(int Num , int min, int max){
  int i;
  for(i=min;i < max;i++){
    
}
*/
 
int main()
{
    long long X, Y;
    int len = 0;
    int i = 1;

  // 整数の入力
    scanf("%lld %lld", &X, &Y);
 // printf("%d\n",N);
  
  while(1){
    long long Xi;
    if(i == 1){
      Xi =X;
    } else {
    Xi = Xi * 2;
    }
//    printf("i:%d X:%lld\n", i, Xi);
    if(Xi > Y){
      break;
    } else {
        len = i;
    }
    i++;
  }
printf("%d\n", len);

/*  
    if((a + b) > (c + d))
    printf("Left\n");
    else if((a + b) == (c + d))
    printf("Balanced\n");
    else
    printf("Right\n");
    return 0;
*/
}