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 Test {
  			static long mod = 1000000007;
  			static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  			public static void main(String[] args) throws IOException  {
  				FastReader sc = new FastReader();
  				int t = sc.nextInt();
  				int arr[][] = new int[(int)2e5+1][19];
  				for( int i = 0 ;i < 19 ; i++) {
  					int temp = 0;
  					for( int j = 1 ; j <= (int)2e5  ; j++) {
  						if( check(1<<i , j)) {
  							temp++;
  						}
  						arr[j][i] = temp;
  					}
  				}
  				while( t-- > 0) {
  					int l = sc.nextInt();
  					int r = sc.nextInt();
  					int ans = Integer.MAX_VALUE;
  //					for( int i = l ; i<= r ; i++) {
  //						out.println(Integer.toBinaryString(i));
  //					}
  					for( int i = 0 ; i<= 18 ; i++) {
  						int temp =0;
  						temp+=arr[r][i] - arr[l-1][i];
  //						for( int j = l ;j <= r ;j++) {
  //							if( check( 1 << i , j)) {
  //								temp++;
  //							}
  //						}
  						if( temp < ans) {
  //							out.println(i);
  							ans = temp;
  						}
  					}
  					out.println(ans);
  				}
  				out.flush();
  			}
  			private static boolean check(int i, int j) {
  				return (i&j) == 0;
  			}
  			/*
  			 * WARNING -> DONT EVER USE ARRAYS.SORT() IN ANY WAY.
  			 * FIRST AND VERY IMP -> READ AND UNDERSTAND THE QUESTION VERY CAREFULLY.
  			 * WARNING -> DONT CODE BULLSHIT , ALWAYS CHECK THE LOGIC ON MULTIPLE TESTCASES AND EDGECASES BEFORE. 
  			 * SECOND -> TRY TO FIND RELEVENT PATTERN SMARTLY.
  			 * WARNING -> IF YOU THINK YOUR SOLUION IS JUST DUMB DONT SUBMIT IT BEFORE RECHECKING ON YOUR END.
  			 */
  			public static boolean ifpowof2(long n ) {
  				return ((n&(n-1)) == 0);
  			}
  
  			static boolean isprime(long x ) {
  				if( x== 2) {
  					return true;
  				}
  				if( x%2 == 0) {
  					return false;
  				}
  				for( long i = 3 ;i*i <= x ;i+=2) {
  					if( x%i == 0) {
  						return false;
  					}	
  				}
  				return true;
  			}
  
  			static boolean[] sieveOfEratosthenes(long n) { 
  				boolean prime[] = new boolean[(int)n + 1];
  				for (int i = 0; i <= n; i++) {
  					prime[i] = true;
  				}
  				for (long p = 2; p * p <= n; p++) {   
  					if (prime[(int)p] == true) {
  						for (long i = p * p; i <= n; i += p)
  							prime[(int)i] = false;
  					}
  				}		 
  				return prime;
  			}
  
  			public static int[] nextLargerElement(int[] arr, int n)	{ 
  				Stack<Integer> stack = new Stack<>();
  				int rtrn[] = new int[n];
  				rtrn[n-1] = -1;
  			stack.push( n-1);
  			for( int i = n-2 ;i >= 0 ; i--){
  			    int temp = arr[i];
  			    int lol = -1;
  			    while( !stack.isEmpty() && arr[stack.peek()] <= temp){
  				if(arr[stack.peek()] == temp ) {
  					lol = stack.peek();
  				}
  				stack.pop();
  			    }
  			    if( stack.isEmpty()){
  				if( lol != -1) {
  					rtrn[i] = lol;
  				}
  				else {
  					rtrn[i] = -1;
  				}
  			    }
  			    else{
  				rtrn[i] = stack.peek();
  			    }
  			    stack.push( i);
  			}
  			return rtrn;
  			}
  
  			static void mysort(int[] arr) {
  				for(int i=0;i<arr.length;i++) {
  			    int rand = (int) (Math.random() * arr.length);
  			    int loc = arr[rand];
  			    arr[rand] = arr[i];
  			    arr[i] = loc;
  			}
  			Arrays.sort(arr);
  		    }
  
  
  			static void mySort(long[] arr) {
  			for(int i=0;i<arr.length;i++) {
  			    int rand = (int) (Math.random() * arr.length);
  			    long loc = arr[rand];
  			    arr[rand] = arr[i];
  			    arr[i] = loc;
  			}
  			Arrays.sort(arr);
  		    }
  
  			static long gcd(long a, long b)
  			{
  				if (a == 0)
  					return b;
  				return gcd(b % a, a);
  			}
  
  
  			 static long lcm(long a, long b)
  			 {
  				 return (a / gcd(a, b)) * b;
  			 }
  
  
  			 static long rightmostsetbit(long n) {
  				 return n&-n;
  			 }
  
  			 static long leftmostsetbit(long n)
  			    {
  				long k = (long)(Math.log(n) / Math.log(2));
  				return 1 << k;
  			    }
  
  			 static HashMap<Long,Long> primefactor( long n){
  				 HashMap<Long ,Long> hm = new HashMap<>();
  				 long temp = 0;
  				 while( n%2 == 0) {
  					 temp++;
  					 n/=2;
  				 }	
  				 if( temp!= 0) {
  					 hm.put( 2L, temp);
  				 }
  				 long c = (long)Math.sqrt(n);
  				 for( long i = 3 ; i <= c ; i+=2) {
  					 temp = 0;
  					 while( n% i == 0) {
  						 temp++;
  						 n/=i;
  					 }
  					 if( temp!= 0) {
  						 hm.put( i, temp);
  					 }
  				 }
  				 if( n!= 1) {
  					 hm.put( n , 1L);
  				 }
  				 return hm;	
  			 }
  
  
  			 static ArrayList<Integer> allfactors(int abs) {
  				 HashMap<Integer,Integer> hm = new HashMap<>();
  				 ArrayList<Integer> rtrn = new ArrayList<>();
  				 for( int i = 2 ;i*i <= abs; i++) {
  					 if( abs% i == 0) {
  						 hm.put( i , 0);
  						 hm.put(abs/i, 0);
  					 }
  				 }
  				 for( int x : hm.keySet()) {
  					 rtrn.add(x);
  				 }
  
  				 if( abs != 0) {
  					 rtrn.add(abs);
  				 }
  
  				 return rtrn;
  			 }
  
  
  			 static class FastReader {
  				 BufferedReader br;
  				 StringTokenizer st;
  
  				 public FastReader()
  				 {
  					 br = new BufferedReader(new InputStreamReader(System.in));
  				 }
  
  				 String next()
  				 {
  					 while (st == null || !st.hasMoreElements()) {
  						 try {
  							 st = new StringTokenizer(br.readLine());
  						 }
  						 catch (IOException e) {
  							 e.printStackTrace();
  						 }
  					 }
  					 return st.nextToken();
  				 }
  
  				 int nextInt() { return Integer.parseInt(next()); }
  
  				 long nextLong() { return Long.parseLong(next()); }
  
  				 double nextDouble()
  				 {
  					 return Double.parseDouble(next());
  				 }	
  
  				 String nextLine()
  				 {
  					 String str = "";
  					 try {
  						 str = br.readLine();
  					 }
  					 catch (IOException e) {
  						 e.printStackTrace();
  					 }
  					 return str;
  				 }
  			 }	
  	}
  
  
  