Minesweeper II

You are standing in position ~0,0~ on an infinite minefield. By making moves down and to the right, you are trying to get to position ~X, Y~.

The "riskiness" of your position is equal to the number of bits in the binary representation of the ~XOR~ between its row and column.

For example, if you're standing at position ~5,17~, then your riskiness is 2, because:

• 5 XOR 17 = 0b101 XOR 0b10001 = 0b10100 = 20.
• There are 2 set bits (1s) in the binary represenation of 20.

The "danger" of a path is defined as the maximum riskiness of all positions along that path.

Given a position ~X, Y~ you want to go to, what is the minimum danger you can have along the journey there?

Input

The first line of each test case contains two integers, ~X~ and ~Y~, which represent the coordinates of the place you want to go to. The values of ~X~ and ~Y~ are constrained by ~0 \leq X, Y \leq 10^6~.

Output

Output the minimum "danger" you can have when travelling from 0,0 toX,Y, by making moves down and to the right.

Example

Input

3 2

Output

2

• Begin at 0,0. Riskiness = bits(0 XOR 0) = 0
• Move to 1,0. Riskiness = bits(1 XOR 0) = 1
• Move to 2,0. Riskiness = bits(2 XOR 0) = 1
• Move to 2,1. Riskiness = bits(2 XOR 1) = 2
• Move to 2,2. Riskiness = bits(2 XOR 2) = 0
• Move to 3,2. Riskiness = bits(3 XOR 2) = 1

The maximum riskiness along the path is 2, so the danger of this path is 2. It can be shown that 2 is the minimum danger possible to get to 3,2.