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 func(a,b):
    bin_a=bin(a)[2:]
    bin_b=bin(b)[2:]
    print(a,b)
    max_len=len(bin_b)
    res1=[]
    b+=1
    for x in range(max_len):
        base=2**(x+1)
        #print(base)
        num=(a//base)*(base//2)+max(a%base-(base//2),0)
#         print(num)
#         print("--------")
        res1.append(num)
    
    res2=[]
    for x in range(max_len):
        base=2**(x+1)
        #print(base)
        num=(b//base)*(base//2)+max(b%base-(base//2),0)
#         print(num)
#         print("--------")
        res2.append(num)
    #print(res1,res2)
    final=[x2-x1 for x1,x2 in zip(res1,res2)]
    #print(final)
    max_num=max(final)
    print(b-a-max_num)

t = int(input())
while t > 0:
    t -= 1
    m, n = map(int, input().split())
    func(m, n)
