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 math
import random
T=int(input())
for t in range(T):
    #arrayz=random.sample(range(1,10000), 2)
    arrayz=[int(i) for i in input().split()]
    l=min(arrayz)
    r=max(arrayz)
    d=dict()
    s=1
    dig=[]
    for i in range(30):
        dig.append(s)
        s*=2
    mul=2
    for i in range(len(dig)):
        start=dig[i]
        if start>r:
            break
        m1=(l-start)/mul
        m1=math.ceil(m1)
        first=m1*mul+start
        if first<=r:
            sets=((r-first+1)//mul)
            tot=sets*(mul)/2
            nw=first+mul*sets
            tot+=max(min(r-nw+1,mul/2),0)
            gap=first-l
            gap-=(mul/2)
            tot+=max(0,gap)    
            d[i]=tot
        else:
            prev=first-mul
            tot=min(r+1,prev+mul/2)-max(prev,l)
            d[i]=tot
        mul*=2

    m=0
    for i in d:
        m=max(m,d[i])
    ans1=int(r-l+1-m)
    print(ans1)
    p='''d=dict()
    for i in range(l,r+1):
        s=bin(i)[2:]
        s=s[::-1]
        for i in range(len(s)):
            if s[i] =='1':
                if i in d:
                    d[i]+=1
                else:
                    d[i]=1
    m=0
    for i in d:
        m=max(m,d[i])
    ans2=(r-l+1-m)
    if ans1!=ans2:
        print("error",ans1,ans2,l,r)
    else:
        pass'''