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.
def find(l ,r):
    if l == r:
        return 0
    big = 0
    while 2**big <= r:
        big += 1
    if 2**big <= l:
        return 0
    biggest = 0
    for i in range(big):
        gap = 2**(i+1)

        q1 = int(l / gap)
        x1 = l % gap
        y1 = int(max(0, x1 - gap / 2))
        z1 = int(gap / 2 * q1 + y1)

        q2 = int((r+1)/gap)
        x2 = (r+1) %gap
        y2 = int(max(0, x2 - gap/2))
        z2 = int(gap/2 * q2 + y2)

        biggest = max(biggest, z2 - z1)

    return r-l+1 - biggest

cases = int(input())
for _ in range(cases):
    line = input().split( )
    l = int(line[0])
    r = int(line[1])
    print(find(l, r))