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
import math
from collections import deque,Counter
from sys import stdin

#sys.setrecursionlimit(10**7)

int1=lambda x: int(x)-1
stinput=lambda :stdin.readline()[:-1]
ii=lambda :int(stinput())
mi=lambda :map(int, stdin.readline().split())
li=lambda :list(mi())
mi1=lambda :map(int1, stdin.readline().split())
li1=lambda :list(mi1())
mis=lambda :map(str, stdin.readline().split())
lis=lambda :list(mis())

from collections import defaultdict
#d=defaultdict(int)
#d=defaultdict(lambda:1)

import heapq
#heapq.heapify(hq)
#heapq.heappop(hq)
#heapq.heappush(hq,1)

mod=10**9+7
Mod=998244353
INF=10**18
ans=0

def f(n):
  cnt=[0]*20
  for i in range(20):
    tmp=0
    m=(n)//(pow(2,i+1))
    tmp+=m*pow(2,i)
    k=m*pow(2,i+1)
    tmp+=max(n-k-pow(2,i)+1,0)
    cnt[i]=tmp
  return cnt

t=ii()
for _ in range(t):
  l,r=mi()
  n=r-l+1
  ans=n
  a=f(r)
  b=f(l-1)
  for i in range(20):
    ans=min(ans,n-(a[i]-b[i]))
  print(ans)