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 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();
	}
}