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.
// package c1615;

import java.io.File;
import java.lang.invoke.MethodHandles;
import java.util.Random;
import java.util.Scanner;

//
// Codeforces Global Round 18 2021-12-24 06:35
// B. And It's Non-Zero
// https://codeforces.com/contest/1615/problem/B
// time limit per test 2 seconds; memory limit per test 256 megabytes
// public class Pseudo for 'Source should satisfy regex [^{}]*public\s+(final)?\s*class\s+(\w+).*'
//
// 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 <a href="https://en.wikipedia.org/wiki/Bitwise_operation#AND">bitwise AND</a> of the
// array non-zero?
//
// A is a binary operation that takes two equal-length binary representations and performs the
// operation on each pair of the corresponding bits.
//
// Input
//
// The first line contains one integer t (1 <=q t <=q 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 <=q l <=q r <=q 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 is 0, as 1\ \& \ 2 = 0. However,
// after deleting 1 (or 2), the array becomes [2] (or [1]), and the 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 is 0. However, after
// deleting 4, 5, and 8, the array becomes [2, 3, 6, 7], and the 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.
//
public class C1615B {
  static final int MOD = (int)1e9+7;
  static final Random RAND = new Random();

  static int[][] zeros = null;

  static void init() {
    if (zeros == null) {
      zeros = new int[200001][20];
      for (int i = 0; i <= 200000; i++) {
        for (int j = 0; j < 20; j++) {
          if ((i & (1 << j)) == 0) {
            zeros[i][j]++;
          }
        }
      }
      for (int i = 1; i <= 200000; i++) {
        for (int j = 0; j < 20; j++) {
          zeros[i][j] += zeros[i-1][j];
        }
      }
    }
  }

  static int solve(int l, int r) {
    // [1,200k]
    init();
    int ans = zeros[r][0] - zeros[l-1][0];
    for (int i = 0; i < 20; i++) {
      ans = Math.min(ans, zeros[r][i] - zeros[l-1][i]);
    }
    return ans;
  }

  static void doTest() {
    long t0 = System.currentTimeMillis();
    int ans = solve(1, 200000);
    System.out.format("%d msec\n", System.currentTimeMillis() - t0);
    System.exit(0);
  }

  public static void main(String[] args) {
    // doTest();
    Scanner in = getInputScanner();
    int T = in.nextInt();
    for (int t = 1; t <= T; t++) {
      int l = in.nextInt();
      int r = in.nextInt();
      int ans = solve(l, r);
      System.out.println(ans);
    }
    in.close();
  }

  static Scanner getInputScanner() {
    try {
      final String USERDIR = System.getProperty("user.dir");
      final String CNAME = MethodHandles.lookup().lookupClass().getSimpleName();
      final File fin = new File(USERDIR + "/io/c" + CNAME.substring(1,5) + "/" + CNAME + ".in");
      return fin.exists() ? new Scanner(fin) : new Scanner(System.in);
    } catch (Exception e) {
      return new Scanner(System.in);
    }
  }
}
