description: 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.
solution: import java.util.*;
  import java.io.*;
  public class s1 {
  	public static FastScanner scan;
  	public static PrintWriter out;
  	public static void main(String[] args) throws Exception {
  		scan=new FastScanner(System.in);
  		out=new PrintWriter(System.out);
   
  //		int T=1;
  	int T=scan.nextInt();
  		BIT[] bit=new BIT[32];
  		int MX=2*(int)1e5;
  		for(int i=0;i<32;i++) bit[i]=new BIT(MX+3);
  		for(int i=1;i<=MX;i++) {
  			for(int j=0;j<32;j++) {
  				if((i&(1<<j))>0) bit[j].update(i,1);
  			}
  		}
  		while(T-->0) {
  			int l=scan.nextInt(),r=scan.nextInt();
  
  			int max=0;
  			for(int i=0;i<32;i++) max=Math.max(max,bit[i].query(r)-bit[i].query(l-1));
  			out.println((r-l+1)-max);
  
  		} out.close();
  	}
  }
  
  class BIT {
  	int[] tree;
  	BIT(int n) {
  		tree=new int[n];
  	}
  	void update(int i,int val) {
  		i++;
  		while(i<tree.length) {
  			tree[i]+=val;
  			i+=i&-i;
  		}
  	}
  
    int query(int i) {
  		int res=0;
  		i++;
  		while(i>0) {
  			res+=tree[i];
  			i-=i&-i;
  		} return res;
  	}
  }
  
  
  
  class Pair {
  	int a,b;
  	Pair(int a,int b) {
  		this.a=a;
  		this.b=b;
  	}
  }
   
  class FastScanner {
  	private InputStream stream;
  	private byte[] buf=new byte[1024];
  	private int curChar;
  	private int numChars;
   
  	public FastScanner(InputStream stream) { this.stream=stream; }
   
  	int read() {
  		if(numChars==-1) throw new InputMismatchException();
  		if(curChar>=numChars) {
  			curChar=0;
  			try { numChars=stream.read(buf); }
  			catch(IOException e) { throw new InputMismatchException(); }
  			if(numChars<=0) return -1;
  		} return buf[curChar++];
  	}
   
  	boolean isSpaceChar(int c) { return c==' '||c=='\n'||c=='\r'||c=='\t'||c==-1; }
   
  	boolean isEndline(int c) { return c=='\n'||c=='\r'||c==-1; }
   
  	int nextInt() { return Integer.parseInt(next()); }
  
  	long nextLong() { return Long.parseLong(next()); }
   
  	double nextDouble() { return Double.parseDouble(next()); }
   
  	String next() {
  		int c=read();
  		while(isSpaceChar(c)) c=read();
  		StringBuilder res=new StringBuilder();
  		do {
  			res.appendCodePoint(c);
  			c=read();
  		} while(!isSpaceChar(c));
  		return res.toString();
  	}
   
  	String nextLine() {
  		int c=read();
  		while(isEndline(c)) c=read();
  		StringBuilder res=new StringBuilder();
  		do {
  			res.appendCodePoint(c);
  			c=read();
  		} while(!isEndline(c));
  		return res.toString();
  	}
  }