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.
T = int(input())
for t in range(T):
    l, r = [int(i) for i in input().split()]
    diff = r-l+1
    i = 0
    ans = diff
    while (1<<i) <= r:
        
        at_r = (r // (2*(1<<i)) )*(1<<i)
        #print('i',i,'at_r',at_r)
        modval = 0
        if i >= 0:
            modval = r % (2*(1<<i))
        #print('modval',modval)
        incr = 0
        if modval >= (1<<i):
            incr = modval - (1<<i) + 1
        total_at_r = at_r + incr
        #print(2**i, 'total',total_at_r)
        #print('-'*50)

        at_l = ( (l-1) // (2*(1<<i)) )*(1<<i)
        modval = 0
        if i >= 0:
            modval = (l-1) % (2*(1<<i))
        #modval -= 1
        incr = 0
        if modval >= (1<<i):
            incr = modval - (1<<i) + 1
        total_at_l = at_l + incr
        #print(total_at_l, total_at_r,'for 2^i=',(1<<i))
        ans = min(ans, diff - (total_at_r - total_at_l))
        i += 1
        
    print(ans)

'''
0
1
2
3
4
5
6
7
8


'''
