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.
from sys import stdin, stdout, maxsize
from math import sqrt, log, factorial as ft, gcd, ceil, floor
from collections import defaultdict

ii = lambda: int(input())
di = lambda: defaultdict(lambda: 0)
dl = lambda: defaultdict(lambda: list())
li = lambda: list(map(int, stdin.readline().strip().split()))
mi = lambda: map(int, stdin.readline().strip().split())
si = lambda: stdin.readline().strip()

"""
px = 1000000
prime = [1] * (px + 1)
for p in range(2, int(sqrt(px)) + 1):
    if prime[p]:
        for i in range(p * p, px + 1, p): prime[i] = 0
prime[0] = prime[1] = 0
"""

for _ in range(ii()):
    l, r = mi()
    if l == r == 1: 
        print(0)
        continue
    ll = []
    l1, l2 = 1, 1
    for i in range(18):
        if l <= l1 <= l2 <= r: ll.append(l1 - l + r - l2)
        elif l1 <= l <= l2 <= r: ll.append(r - l2)
        elif l1 <= l <= r <= l2: ll.append(0)
        elif l <= l1 <= r <= l2: ll.append(l1 - l)
        l1 += l1
        l2 += l1
    ans = min(ll)
    k = (r - l) // 2
    if r % 2 or l % 2: k += 1
    ans = min(ans, r - l + 1 - k)
    print(ans)
        