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 <bits/stdc++.h>
using namespace std;

#define pq priority_queue
#define mp make_pair
#define cauto const auto &

#define REP(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define all(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())

#define clr(ar, val) memset(ar, val, sizeof(ar))

typedef long long int64;
typedef long long lint;
typedef long long lli;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;

typedef pair<int, int> pint;
typedef pair<long, long> plong;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;
typedef pair<ll, ll> pll;

const int iinf = 1 << 29;
const long long linf = 1ll << 61;

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 = ll> T in() {
    T x;
    cin >> x;
    return (x);
}
template <class T> void print(T &x) {
    cout << x << '\n';
    return;
}

template <class T> ll sized_subset(T &comb) {
    T x = comb & -comb, y = x + comb;
    return ((comb & ~y) / x) >> 1 | y;
}

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

int main() {
    ll x=in();
    ll y=in();
    int ans=0;
    while(y>=x){y/=2;ans++;}
    print(ans);
    return 0;
}