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

public class Codeforces {

  public static void main(String[] args) throws Exception {
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter writer = new PrintWriter(System.out);
      int T = Integer.parseInt(br.readLine());
      int[][] bits = new int[200010][20];
      for(int i = 1; i < 200001; i++) {
        int x = i;
        int j = 0;
        while(x != 0) {
          bits[i][j] = x%2;
          x /= 2;
          j++;
        }
      }
      for(int i = 0; i < 20; i++) {
        for(int j = 1; j < 200001; j++) {
          bits[j][i] += bits[j-1][i];
        }
      }
      while(T-- > 0) {
        StringTokenizer st = new StringTokenizer(br.readLine());
        int l = Integer.parseInt(st.nextToken());
        int r = Integer.parseInt(st.nextToken());
        int ans = 0;
        for(int i = 0; i < 20; i++) {
          int count = bits[r][i]-bits[l-1][i];
          ans = Math.max(ans, count);
        }
        writer.println(r-l-ans+1);
      }
      writer.close();
      br.close();
  }

}
