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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
 
public class A {
	
public static void main(String[] args) {
		
		FastScanner sc = new FastScanner();
		int t = sc.nextInt();
		for(int tt = 0 ; tt < t; tt++) {
			int l = sc.nextInt(), r = sc.nextInt();
			int mid = (r+1)/2;
			if(mid >= l) {
				System.out.println(mid - 1);
			}
			else {
				System.out.println(r%l);
			}
		}
	}
	
	public static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}

  
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }
 
    public static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
    public static long pow(long base, long exp, long MOD) {
        base %= MOD;
        long ret = 1;
        while (exp > 0) {
            if ((exp & 1) == 1) ret = ret * base % MOD;
            base = base * base % MOD;
            exp >>= 1;
        }
        return ret;
    }
 
 
    public static long max(long... ar) {
        long ret = ar[0];
        for (long itr : ar) ret = Math.max(ret, itr);
        return ret;
    }
 
    public static int max(int... ar) {
        int ret = ar[0];
        for (int itr : ar) ret = Math.max(ret, itr);
        return ret;
    }
 
    public static long min(long... ar) {
        long ret = ar[0];
        for (long itr : ar) ret = Math.min(ret, itr);
        return ret;
    }
 
    public static int min(int... ar) {
        int ret = ar[0];
        for (int itr : ar) ret = Math.min(ret, itr);
        return ret;
    }
 
 
    public static long sum(long... ar) {
        long sum = 0;
        for (long itr : ar) sum += itr;
        return sum;
    }
 
    public static long sum(int... ar) {
        long sum = 0;
        for (int itr : ar) sum += itr;
        return sum;
    }
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		
		public String next() {
			while (!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());
		}
		
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
	}
 
}


 
 
 
 
 
 
 
 
