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 os
import sys
from io import BytesIO, IOBase

from collections import defaultdict, Counter
from typing import AsyncIterator

input = lambda: sys.stdin.readline().rstrip("\r\n")
import math

def I():
    return(int(input()))
def lint():
    return(list(map(int,input().split())))
def insr():
    s = input().strip()
    return(list(s[:len(s)]))
def invr():
    return(map(int,input().split()))
# Python program to print all
# primes smaller than or equal to
# n using Sieve of Eratosthenes

import heapq

def ask(a,b,c):
    print(f"? {a} {b} {c}")
    sys.stdout.flush()
    res=int(input())
    sys.stdout.flush()
    return res



t=  I()
for _ in range(t):
    l,r=lint()
    tmp=int(math.log(r,2))
    tmp=2**tmp
    ans=0
  
    ans=max(ans,max(r-max(tmp,l)+1,tmp-max(tmp//2,l)))
    
    if l%2==0:
        ans=max(ans,(r-l+1)//2)
    else:
        ans=max(ans,(r-l+2)//2)
    print((r-l+1)-ans)


    