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 B {
    static MyScanner sc;
    static PrintWriter out;
    static int[][] pref;
    static {
        sc = new MyScanner();
        out = new PrintWriter(System.out);
        int n = (int)2e5;
        pref = new int[n + 1][30];
        for(int j = 0; j < 30; j++) {
            for(int i = 1; i <= n; i++) {
                pref[i][j] = pref[i - 1][j] + ((i >> j) & 1);
            }
        }
    }
    public static void solve() {
        int l = sc.nextInt();
        int r = sc.nextInt();
        int ans = 0;
        for(int i = 0; i < 30; i++)
            ans = Math.max(ans, pref[r][i] - pref[l - 1][i]);
        out.println(r - l + 1 - ans);
    }
    public static void main(String[] args) {
        int t = sc.nextInt();
        while(t-- > 0)
            solve();
        out.flush();
    }
}

class MyScanner {
    BufferedReader br;
    StringTokenizer tok;
    MyScanner() {
        try { br = new BufferedReader(new InputStreamReader(System.in)); }
        catch(Exception e) { System.out.println(e); }
        tok = new StringTokenizer("");
    }
    public String next() {
        try {
            while(!tok.hasMoreTokens()) tok = new StringTokenizer(br.readLine());
        }
        catch(Exception e) { System.out.println(e); }
        return tok.nextToken();
    }
    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong() {
        return Long.parseLong(next());
    }
    public double nextDouble() {
        return Double.parseDouble(next());
    }
}
