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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define int long long
using ll=long long;
using vi=vector<int>;
using pii=pair<int,int>;
#define ALL(c) begin(c),end(c)
#define RALL(c) rbegin(c),rend(c)
#define ITR(i,b,e) for(auto i=(b);i!=(e);++i)
#define FORE(x,c) for(auto &x:c)
#define REPF(i,a,n) for(int i=a,i##len=(int)(n);i<i##len;++i)
#define REP(i,n) REPF(i,0,n)
#define REPR(i,n) for(int i=(int)(n);i>=0;--i)
#define SZ(c) ((int)c.size())
#define CONTAIN(c,x) (c.find(x)!=end(c))
#define OUTOFRANGE(y,x,h,w) (y<0||x<0||y>=h||x>=w)
#define dump(...)
const int DX[9]={0,1,0,-1,1,1,-1,-1,0},DY[9]={-1,0,1,0,-1,1,1,-1,0};
#define INF (1001001001)
#define INFLL (1001001001001001001ll)
template<class T> ostream& operator << (ostream &os,const vector<T> &v) {
    ITR(i,begin(v),end(v)) os<<*i<<(i==end(v)-1?"":" "); return os; }
template<class T> istream& operator >> (istream &is,vector<T> &v) {
    ITR(i,begin(v),end(v)) is>>*i; return is; }
template<class T> istream& operator >> (istream &is, pair<T,T> &p) {
    is>>p.first>>p.second; return is; }
template <class T> bool chmax(T &a,const T &b) {if(a<b) {a=b;return 1;} return 0;}
template <class T> bool chmin(T &a,const T &b) {if(b<a) {a=b;return 1;} return 0;}
template <class T> using heap = priority_queue<T,vector<T>,greater<T>>;
struct before_main_function {
    before_main_function() {
        #ifdef int
            #undef INF
            #define INF INFLL
            #define stoi stoll
        #endif
        cin.tie(0);ios::sync_with_stdio(false);
        cout<<setprecision(15)<<fixed;
        #define endl "\n"
    }
} before_main_function;
//------------------8<------------------------------------8<--------------------

const double EPS=1e-6;
signed main() {
    int X,Y;
    cin>>X>>Y;
    int ans=0;
    REP(i,63) {
        if(log2(Y)>i+log2(X) || abs(log2(Y)-log2(X)-i)<EPS) {
            ans=i+1;
        }
    }
    cout<<ans<<endl;
    return 0;
}