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.
a=[(bin(i))[2:] for i in range(1,200000+1)]
for i in range(200000):
    a[i]=a[i].rjust(20,'0')
x=[0 for i in range(20)]
sol=[]
sol.append(list(x))
# print(x)
# print(sol[0])
for i in range(200000):
    for j in range(20):
        if a[i][j]=='0':
            x[j]+=1
    # if i==200000-2:
    #     print(sol[0])
    sol.append(list(x))
    # if i==200000-2:
    #     print(sol[0])
# print("hello",sol[200000])
for jksdfh in range(int(input())):
    l,r=list(map(int,input().split()))
    if l==r:
        print(0)
    l=sol[l-1]
    r=sol[r]
    ans=[]
    for i in range(20):
        ans.append(r[i]-l[i])
    print(min(ans))