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.
import sys

def readnum():
    input = sys.stdin.readline
    a=list(map(int,input().split()))
    return a

def printf(val):
    sys.stdout.write(str(val) + '\n')

def power(a, b, mod):
    if b == 1:
        return a
    if b == 0:
        return 1

    ans = power(a, b//2, mod)
    ans *= ans

    if b % 2:
        ans *= a
    
    ans = ans % mod
    return ans


def factorial(n, start=1, skip=None):
    res = start
    for i in range(1, n+1):
        if i == skip:
            continue
        res *= i
#        res = res % mod

    return res

def make_deg(l):
    deg = {}
    for i in l:
        if not i in deg:
            deg[i] = 0
        deg[i] += 1
    return deg

def get_num_of_bits(n):
    res = []
    for i in range(1, 31):
        res.append( 2**(i-1)*(n//(2**i)) + max(0, (n % (2**i)) - 2**(i-1)))
    return res

def minus(l, r):
    res = []
    for i in range(len(l)):
        res.append(r[i] - l[i])
    return res

for _ in range(readnum()[0]):
    l, r = readnum()
#    print (l, get_num_of_bits(l+1))
#    print (r, get_num_of_bits(r+1))
    res = minus(get_num_of_bits(l), get_num_of_bits(r+1))
#    print (res)
    print (r - l - max(res) + 1)
#    print ("" )
