Funny Bits

Problem Statement

Let ~a~ be the average (rounded down) of all positive integers up to ~k~-bits with ~n~ bits set.

(A bit is "set" if it equals one. For example, in the binary number 1101, there is a total of 3 bits set.)

Given ~k~ and ~a~, find ~n~.

Input

Your only line of input will contain two space-separated integers, ~k~ and ~a~.

Output

One line, containing only the number ~n~.

Constraints

For all test cases:

• ~1 \le k \le 30~
• ~1 \le a \le 10^{10}~
• There will always be a unique valid ~n~ for the given ~k~ and ~a~.

Sample Test Cases

Example 1

Input

4 7

Output

2

Explanation

If we let ~n = 2~, then there are a total of ~6~ integers (3, 5, 6, 9, 10, 12) that can be represented with at most ~k = 4~ bits with exactly ~n = 2~ bits set:

3 = 0011
5 = 0101
6 = 0110
9 = 1001
10 = 1010
12 = 1100

The average of these numbers is ~7.5~, which rounds down to ~7~. Thus, ~n = 2~.

Example 2

Input

3 2

Output

1

Explanation

If we let ~n = 1~, then there are a total of ~3~ integers (1, 2, 4) that can be represented with at most ~k = 3~ bits with exactly ~n = 1~ bit set:

1 = 001
2 = 010
4 = 100

The average of these numbers is ~2.33~, which rounds down to ~2~. Thus, ~n = 2~.

Example 3

Input

7 90

Output

5

Explanation

If we let ~n = 5~, then there are a total of ~21~ integers that can be represented with at most ~k = 7~ bits with exactly ~n = 5~ bits set. These numbers sum to ~1905~ and averages to ~90.71~, which rounds down to ~90~. Thus, ~n = 5~.