You are given two integers l and r, l≤ r. Find the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.

As a reminder, a mod b is a remainder we get when dividing a by b. For example, 26 mod 8 = 2.

Input

Each test contains multiple test cases.

The first line contains one positive integer t (1≤ t≤ 10^4), denoting the number of test cases. Description of the test cases follows.

The only line of each test case contains two integers l, r (1≤ l ≤ r ≤ 10^9).

Output

For every test case, output the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.

Example

Input


4
1 1
999999999 1000000000
8 26
1 999999999


Output


0
1
12
499999999

Note

In the first test case, the only allowed pair is (a, b) = (1, 1), for which a mod b = 1 mod 1 = 0.

In the second test case, the optimal choice is pair (a, b) = (1000000000, 999999999), for which a mod b = 1.
import java.util.*;
public class B {
	public static boolean[] sieveOfEartosthenes(int n) {
		boolean isPrime[] = new boolean[n+1];
		Arrays.fill(isPrime, true);
		isPrime[0]=false;
		isPrime[1]=false;
		
		for(int i=2;i*i<n;i++) {
			for(int j=2*i;j<n;j+=i) {
				isPrime[j]=false;
			}
		}
		
		return isPrime;
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t=sc.nextInt();
		boolean[] arr=sieveOfEartosthenes(100);
		while(t-->0) {
			int n=sc.nextInt();
			String s=sc.next();
			
			char[] ch = s.toCharArray();
			
			boolean flag=false;
			
			for(int i=0;i<n;i++) {
				if(ch[i]=='1' || ch[i]=='4' || ch[i]=='6' || ch[i]=='8' || ch[i]=='9') {
					System.out.println(1);
					System.out.println(ch[i]);
					flag=true;
					break;
				}
			}
			
			if(flag==false) {
				
				for(int i=0;i<n;i++) {
					int temp1=Integer.parseInt(Character.toString(ch[i]));
					temp1*=10;
					for(int j=i+1;j<n;j++) {
						int temp2=Integer.parseInt(Character.toString(ch[j]));
						int num=temp1+temp2;
						if(arr[num]==false) {
							System.out.println(2);
							System.out.println(num);
							break;
						}
					}
				}
				
			}
			
			
			
		}		
	}
}
