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)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#pragma warning(disable : 4996)
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>
#include<bitset>
#include<limits.h>
#include<cassert>
#include<iterator>
#include<complex>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<time.h>
#include<random>
#include<array>


using namespace std;

#define REP(i, n) for(int i = 0;i < n;++i)
#define REPR(i, n) for(int i = n-1;i >= 0;--i)
#define FOR(i, m, n) for(int i = m;i < n;++i)
#define FORR(i, m, n) for(int i = m-1;i >= n;--i)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v,n) reverse(v,v+n);
#define VREVERSE(v) reverse(v.begin(), v.end())
#define ll long long
#define print(x) cout<<(x)<<'\n'
#define pe(x) cout<<(x)<<" "
#define DEBUG(x) cout<<#x<<": "<<x<<endl
#define lb(v,n) lower_bound(v.begin(), v.end(), (n))
#define ub(v,n) upper_bound(v.begin(), v.end(), (n))
#define int long long
//#define double long double
#define all(x) (x).begin(), (x).end()
#define print_space(v) REP(i,v.size())cout << v[i] << ((i == v.size() - 1) ? "\n" : " ")
template<typename T1, typename T2> inline void chmin(T1 & a, T2 b) { if (a > b) a = b; }
template<typename T1, typename T2> inline void chmax(T1& a, T2 b) { if (a < b) a = b; }
typedef pair<int, int>P;
typedef array<int, 3> arr3;
std::random_device rd;
std::mt19937 mt(rd());
constexpr ll MOD = 998244353; constexpr int MAX = 2000020;
const double pi = acos(-1); constexpr double EPS = 1e-8;
constexpr ll INF = 1e18;
void y(bool cond) {
	if (cond)print("Yes");
	else print("No");
}

void solve() {
	ll X, Y; cin >> X >> Y;
	int ans = 0;
	int now = X;
	while (now <= Y) {
		now *= 2;
		ans++;
	}
	print(ans);
}


signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	//int q; cin >> q;
	//while (q--)
	solve();
}