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 math import ceil, log2, floor,sqrt,gcd
from collections import defaultdict,deque
from bisect import bisect_left
import sys
import os.path
from itertools import permutations
sys.setrecursionlimit(200000)
if (os.path.exists('input.txt')):
    sys.stdin = open("input.txt", "r+")
    sys.stdout = open("output.txt", "w")


# input = sys.stdin.readline
# print=sys.stdout.write
def MI():
    return map(int, input().split())

mod=10**9+7




def solve(testcase):
    # input()
    # n = int(input())
    # a=list(MI())
    l,r=MI()
    c1=0
    l1=l
    r1=r
    while(l>1):
        l=l//2
        c1+=1
    c2=0
    while(r>1):
        r=r//2
        c2+=1
    # print(c1,c2)
    l=l1
    r=r1
    if c1==c2:
        print(0)
    else:
        ans=10**9
        # print(r+1,2**(c1+1))
        ans=min(ceil((r+1-2**(c1+1))),ans)
        # print(ans)

        ans=min(ans, ((2**c2) - l))

        # ans= min(  ans,  )
        k= r+1-2**c2 
        c2-=1
        k+=2**c2 -l
        ans=min(ans,k)
        an=ceil((r-l)/2)
        print(min(ans,an))

    

    
    
    
t=int(input())
# t=1
for _ in range(t):
    solve(_+1)

