You are given an array consisting of all integers from [l, r] inclusive. For example, if l = 2 and r = 5, the array would be [2, 3, 4, 5]. What's the minimum number of elements you can delete to make the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the array non-zero?

A bitwise AND is a binary operation that takes two equal-length binary representations and performs the AND operation on each pair of the corresponding bits.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.

The first line of each test case contains two integers l and r (1 ≤ l ≤ r ≤ 2 ⋅ 10^5) — the description of the array.

Output

For each test case, output a single integer — the answer to the problem.

Example

Input


5
1 2
2 8
4 5
1 5
100000 200000


Output


1
3
0
2
31072

Note

In the first test case, the array is [1, 2]. Currently, the bitwise AND is 0, as 1\ \& \ 2 = 0. However, after deleting 1 (or 2), the array becomes [2] (or [1]), and the bitwise AND becomes 2 (or 1). This can be proven to be the optimal, so the answer is 1.

In the second test case, the array is [2, 3, 4, 5, 6, 7, 8]. Currently, the bitwise AND is 0. However, after deleting 4, 5, and 8, the array becomes [2, 3, 6, 7], and the bitwise AND becomes 2. This can be proven to be the optimal, so the answer is 3. Note that there may be other ways to delete 3 elements.
import math
problem_num = int(input())
for problem in range(problem_num):
    arr = list(map(int,input().split()))
    low,high = arr
    num_0_lst = []
    num_1_lst = []
    ran = int(math.log2(high))
    for i in range(0,ran + 1):
        x = high//2**(i+1) * 2**(i+1) - 1
        y = low // 2 ** (i+1) * 2 ** (i+1)

        length = x - y + 1
        count_0 = length // 2
        count_1 = length // 2

        count_to_high = high - x
        count_to_low = low - y
        if count_to_high // 2 ** i >= 1:
            count_0 += 2 ** i
            count_1 += count_to_high % 2 ** i
        else:
            count_0 += count_to_high

        if count_to_low // 2 ** i >= 1:
            count_0 -= 2 ** i
            count_1 -= count_to_low % 2 ** i
        else:
            count_0 -= count_to_low
        num_0_lst.append(count_0)
        num_1_lst.append(count_1)

    # print(min(num_0_lst + num_1_lst))
    print(min(num_0_lst))



