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.Arrays;
import java.util.StringTokenizer;

public class Solution {
    static class Reader{
        private final BufferedReader reader;
        private StringTokenizer tokenizer;
        Reader(InputStream input)
        {
            this.reader = new BufferedReader(new InputStreamReader(input));
            this.tokenizer = new StringTokenizer("");
        }
        public String next() throws IOException {
            while(!tokenizer.hasMoreTokens())
            {
                tokenizer = new StringTokenizer(reader.readLine());
            }
            return tokenizer.nextToken();
        }
        public int nextInt() throws IOException
        {
            return Integer.parseInt(next());
        }
        public double nextDouble() throws IOException
        {
            return Double.parseDouble(next());
        }
        public long nextLong() throws IOException
        {
            return Long.parseLong(next());
        }

        public int[] nextIntArr(int n) throws IOException {
            int[] arr = new int[n];
            for(int i=0;i<n;i++)arr[i]=nextInt();
            return arr;
        }
        public long[] nextLongArr(int n)throws IOException {
            long[] arr = new long[n];
            for(int i=0;i<n;i++)arr[i]=nextLong();
            return arr;
        }
        public char[] nextCharArr(int n) throws IOException {
            char[] arr = new char[n];
            for(int i=0;i<n;i++)arr[i] = next().charAt(0);
            return arr;
        }
    }
    public static void main(String[] args) throws IOException {
        Reader scan = new Reader(System.in);
        StringBuffer sb = new StringBuffer();
        BufferedOutputStream b = new BufferedOutputStream(System.out);
        int t = scan.nextInt();
        int[][] arr = new int[32][200001];
        for(int i=0;i<32;i++)
        {
            int pow2 = 1<<i;
            for(int j=1;j<200001;j++)
            {
                if((j&pow2) == pow2)arr[i][j] = arr[i][j-1] + 1;
                else arr[i][j] = arr[i][j-1];
            }
        }
        while(t-->0)
        {
            int l = scan.nextInt();
            int r = scan.nextInt();
            int min = Integer.MAX_VALUE;
            for(int i=0;i<32;i++)
            {
                int x = arr[i][r] - arr[i][l-1] ;
                min = Math.min(r-l+1 - x, min);
            }
            sb.append(min+"\n");
        }
        b.write((sb.toString()).getBytes());
        b.flush();
    }




}
