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 B {
   public static void main(String[] hi) {
	  int[][] dp = new int[200001][32];
	  int[] ones = new int[32];
	  for (int i=0; i<dp.length; i++) {

		  String s = Integer.toBinaryString(i);
		  int digit = 1;
		  for (int k=s.length()-1; k>=0; k--) {
			  if (s.charAt(k) == '1') {
				  //System.out.println(ones[32-digit]);
				  ones[32-digit] += 1;
			  }
			  digit++;
		  }
		  int[] copy = new int[ones.length];
		  for (int k=0; k<ones.length; k++) {
			  copy[k] = ones[k];
		  }
		  dp[i] = copy;
		  //System.out.println(Arrays.toString(ones));
	  }
	  FastIO io = new FastIO();
      int t = io.nextInt();
      while (t --> 0) {
         int l = io.nextInt(), r = io.nextInt();         
         int[] array = new int[32];
         int max = 0;
         //System.out.println(Arrays.toString(dp[r]));
         //System.out.println(Arrays.toString(dp[l-1]));
         for (int i=0; i<array.length; i++) {
        	 array[i] = dp[r][i]-dp[l-1][i];
        	 
        	 max = Math.max(max, array[i]);
         }
         //System.out.println(Arrays.toString(array));
         io.println((r-l+1)-max);
      }
      io.close();
   }
   static class FastIO extends PrintWriter {
      BufferedReader br;
      StringTokenizer st;
      public FastIO() {
         super(System.out);
         br = new BufferedReader(new InputStreamReader(System.in));
      }
      public String next() {
         try {
            while (st == null || !st.hasMoreTokens()) {
               st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
         } 
         catch (Exception e) {
            return null;
         }
      }
      public int nextInt() {
         return Integer.parseInt(next());
      }
      public double nextDouble() {
         return Double.parseDouble(next());
      }
      public long nextLong() {
         return Long.parseLong(next());
      }
      public void close() {
         try {
            br.close();
         } 
         catch (IOException e) {
            e.printStackTrace();
         }
         super.close();
      }
   }
   static class input extends PrintWriter {
      BufferedReader br;
      StringTokenizer st;
      public input() {
         super(System.out);
         br = new BufferedReader(new InputStreamReader(System.in));
      }
      public String next() {
         try {
            while (st == null || !st.hasMoreTokens()) {
               st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
         } 
         catch (Exception e) {
            return null;
         }
      }
      public void close() {
         try {
            br.close();
         } 
         catch (IOException e) {
            e.printStackTrace();
         }
         super.close();
      }
   }
}