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 java.io.*;
import java.util.*;



public class Test
{
    final static FastReader fr = new FastReader();
    final static PrintWriter out = new PrintWriter(System.out) ;
    static int mod = 1000000007;
    static long dp[] ;

    static long fastpower(long a, long b)
    {
        long ans = 1 ;
        while (b > 0)
        {
            if ((b & 1) != 0) ans *= a ;
            a *= a  ;
            b >>= 1 ;
        }
        return ans ;
    }

    static void solve()
    {
        long l = fr.nextInt();
        long r = fr.nextInt();

        long n = r ;

        long p = -1 ;
        while (n != 0)
        {
            n /= 2 ;
            p++ ;
        }

        long a = r - fastpower(2, p)+1 ;
        long b  = fastpower(2,p) - fastpower(2, p - 1) ;

        long totalNum = r - l + 1 ;
        long ans ;
        if (totalNum%2 == 0) {
            ans = totalNum/2 ;
        } else ans = totalNum/2 + 1 ;
        
        ans = Math.max(ans, Math.max(a, b)) ;

        out.println(totalNum - ans);




    }

    static int getBit(int n , int pos)
    {
        int mask = 1 << pos ;

        if ((n&mask) == 0) return 0 ;
        return 1 ;

    }

    public static void main(String[] args)
    {
        int t =  fr.nextInt();
        while (t-- > 0)
        {
            solve() ;
        }
        out.close() ;
    }

    static long gcd(long a, long b)
    {
        if (b == 0) return a ;
        return gcd(b,  a%b) ;
    }


    static boolean isVowel(char ch)
    {
        if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u')
            return true ;
        return false ;
    }

    static int lower_bound(List<Integer> arr, int key)
    {
        int pos = Collections.binarySearch(arr, key) ;
        if (pos < 0)
        {
            pos = -(pos + 1) ;
        }
        return pos ;
    }

    static int upper_bound(List<Long> arr, long key)
    {
        int pos = Collections.binarySearch(arr, key);
        pos++ ;
        if (pos < 0)
        {
            pos = -(pos) ;
        }
        return pos ;
    }


    static class FastReader {

        BufferedReader br;

        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(
                    new InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() { return Integer.parseInt(next()); }

        long nextLong() { return Long.parseLong(next()); }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }


}