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.
from collections import Counter
from queue import Queue
import math


def func2(m, n):
    if m < (1 << n):
        return 0
    if m == (1 << n):
        return 1

    if m & (1 << n) != 0:
        k = (m | ((1 << n) - 1)) - ((1 << n) - 1)
        if k == (1 << n):
            return (m - k + 1)
        else:
            M = (m - k + 1) + (k - (1 << n)) // 2
            return M
    else:
        for k in range(n + 1, m):
            if m & (1 << k) != 0:
                kk = (m | ((1 << k) - 1)) - (1 << k)
                if kk >= m:
                    raise Exception
                return func2(kk, n)


if __name__ == '__main__':
    T = int(input())

    for t in range(T):
        l, r = [int(a) for a in input().split()]

        ans = r
        for n in range(r):
            if (1 << n) > r:
                break
            ll, rr = func2(l - 1, n), func2(r, n)
            #print(ll, rr)
            ans = min(ans, (r - l + 1) - (rr - ll))
        print(ans)
