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.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define int long long
#define ft first
#define sc second
using namespace std;
const int mod=1e9+7,INF=1e17;

main(){
   int t;
   cin >> t;
   while(t--){
   	int l,r,ans=0,tot=INF;
   	cin >> l >> r;
   	int dist=r-l+1;
   	int p=1;
   	while(p*2<=l){
   		p*=2;
	   }
	   int c=p;
	   while(c*2<=r){
	   	c*=2;
	   }
	   int res=INF;
	   if(p<l){
	   	res=min(l-(l-p),res);
	   	p*=2;
	   }
	   for(int j=p;j<=c;j++){
	   	if((r-j+2)%(j*2)==0){
	   		res=min(res,dist-((r-j+2)/(j*2)));
		   }
	   }
	   while(p*2<=c){
	   	res=min(dist-p,res);
	   	p*=2;
	   }
	   res=min(res,dist-(r-c+1));
   	cout << res << endl;
   }
}
