solution

貪欲な感じ。列$A = (X, 2X, 4X, \dots, 2^{k-1}X)$で$Y \lt 2^kX$なものを作ればよい。$O(\log Y)$。

implementation

#!/usr/bin/env python3
x, y = map(int, input().split())
cnt = 0
while x <= y:
    cnt += 1
    x *= 2
print(cnt)