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.
t=int(input())
for _ in range(t):
    l,r=input().split(" ")
    l=int(l)
    r=int(r)
    length=r-l+1
    l-=1
    a=str(bin(int(l)))[2:]
    b=str(bin(int(r)))[2:]
    temp=len(b)-len(a)
    a='0'*temp+a
    i=2
    ans=0
    while(r*2//i!=0):
        temp1=(l//i)*i/2
        if(l*2//i==0):
            temp1=0
        if(int(temp1)!=temp1):
            temp1=int(temp1+0.5)
        temp2=(r//i)*i/2
        if(int(temp2)!=temp2):
            temp2=int(temp2+0.5)
        #print(i,temp1,temp2)
        if(r%i!=0 and r>i):
            temp=r-((r//i)*i+i/2)+1
            if(temp>0):
                temp2+=temp
        if(l%i!=0 and l>i):
            temp=l-((l//i)*i+i/2)+1
            if(temp>0):
                temp1+=temp
        tep=temp2-temp1
        if(tep>ans):
            ans=tep
        i=i*2
    print(int(length-ans))


