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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<vector>
#include<queue>
#include<map>
#define rep(index,num) for(int index=0;index<num;index++)
#define rep1(index,num) for(int index=1;index<=num;index++)
#define scan(argument) cin>>argument
#define prin(argument) cout<<argument<<endl
#define kaigyo cout<<endl
#define eps 1e-15
#define mp(a1,a2) make_pair(a1,a2)
typedef long long ll;
using namespace std;
typedef pair<ll,ll> pll;
typedef pair<int,int> pint;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<pint> vpint;
typedef vector<pll> vpll;
ll INFl=1e+18+1;
int INF=1e+9+1;
int main(){
	ll X,Y;
	scan(X);scan(Y);
	int count=1;
	while(1){
		if((X<<1) >Y) break;
		else{
			X<<=1;
			count++;
		}
	}
	prin(count);
	return 0;
}