Given a number N and a bit number K, check if K^th bit of N is set or not. A bit is called set if it is 1. Position of set bit '1' should be indexed starting with 0 from LSB side in binary representation of the number. Consider N = 4(100):  0th bit = 0, 1st bit = 0, 2nd bit = 1.

Input:
The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contain an integer  N. The second line of each test case contains an integer  K.

Output:
Corresponding to each test case, print "Yes" (without quotes) if K^th  bit is set else print "No" (without quotes) in a new line.

Constraints:
1 ≤ T ≤ 200
1 ≤ N ≤ 10⁹
0 ≤ K ≤ floor(log₂(N) + 1)

Example:
Input:
3
4
0
4
2
500
3

Output:
No
Yes
No

Explanation:
Testcase 1: Binary representation of 4 is 100, in which 0^th bit from LSB is not set. So, answer is No.
 

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