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 nonZero(arr):
    l = len(bin(arr[1]))-2
    bit_count = [0]*(l)
    for i in range(l):
        zero = 0
        r = (arr[1] - arr[0])//(2**(i+1))
        zero += r*(2**i)
        idx1 = arr[0]%(2**(i+1))
        val1 = idx1//(2**(i))
        idx2 = arr[1]%(2**(i+1))
        val2 = idx2//(2**(i))
        # if i == l - 2:
        #     print(2**i)
        #     print(idx1, val1, idx2, val2)
        if val1 == 0 and val2 == 1:
            zero += (2**i - idx1)
        elif val2 == 0 and val1 == 1:
            zero += (idx2 + 1)
        elif val2 == val1:
            if val2 == 1:
                if arr[1] - arr[0] > 2**i:
                    zero += 2**i
            else:
                if arr[1] - arr[0] > 2**i:
                    zero += 2**i - idx1 + idx2 + 1
                else:
                    zero += arr[1] - arr[0] + 1
        bit_count[i] = zero
    return min(bit_count)


r = int(input())
res = []
for i in range(r):

    arr = list(map(int, input().split()))
    res.append(nonZero(arr))
    
for i in res:
    print(i)