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 math import gcd,sqrt,log2,ceil,floor
import heapq as hq
from bisect import bisect_right, bisect_left, bisect
import itertools

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

from sys import stdin, stdout, setrecursionlimit
def isPowerOfTwo(x):
    return (x and (not(x & (x - 1))))
def countOdd(L, R):
 
    N = (R - L) // 2
 
    # if either R or L is odd
    if (R % 2 != 0 or L % 2 != 0):
        N += 1
    return N
MOD = int(1e9) + 7
alp = 'abcdefghijklmnopqrstuvwxyz'
v = {'a':1, 'e':1, 'i':1, 'o':1, 'u':1}
c = {}
T = 1
T = int(stdin.readline())
for _ in range(T):
    # n = int(stdin.readline())
    # b = list(map(int, stdin.readline().rstrip().split()))
    l,r = list(map(int, stdin.readline().rstrip().split()))
    # digits = list(stdin.readline().strip('\n'))
    # a = list(stdin.readline().strip('\n'))
    # b = list(stdin.readline().strip('\n'))
    # t = str(stdin.readline().strip('\n'))
    # print(log2(l), log2(r))
    L = ceil(log2(l))
    R = floor(log2(r))
    # print(L,R)
    ans = max(2**L - l + 1, r - 2**(R) + 1, 2**(R-1), countOdd(l,r))
    print(r-l+1-ans)