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 bisect import bisect_left, bisect
from sys import stdin, stdout
rd = lambda: list(map(int, stdin.readline().split()))
rds = lambda: stdin.readline().rstrip()
ii = lambda: int(stdin.readline())
INF = 1 << 62
mod = 10**9 + 7

ps = [1]
i = 2
while i <= 400000:
    ps.append(i)
    i *= 2

for _ in range(ii()):
    l, r = rd()
    if l == r:
        print(0)
        continue

    i = bisect_left(ps, l)
    ll = ps[i]
    if ll > r:
        print(0)
        continue

    to_del = ll - l
    x = ll
    # for each 2**(n+i) <= r remove 2**(n+1) + x,
    # x = 0 .. 2**n-1
    i += 1
    while ps[i] <= r:
        d = min(x, r-ps[i]+1)
        #print('i', i, 'd', d)
        to_del += d
        i += 1
    
    #print('to_del', to_del, 'll-l', ll-l)
    print(to_del)


#stdout.write(' '.join(map(str, ar)))
#stdout.write(f'{res}')

