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.
/*Everything is Hard 
 * Before Easy 
 * Jai Mata Dii 
 */ 
 
import java.util.*;
import java.io.*; 
  
public class Main {
	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; }} 
	static long mod = (long)(1e9+7); 
	// static long mod = 998244353; 
//	 static Scanner sc = new Scanner(System.in); 
	static FastReader sc = new FastReader(); 
	static PrintWriter out = new PrintWriter(System.out);
	public static void main (String[] args) { 
		int ttt = 1;
		ttt = sc.nextInt();
		z :for(int tc=1;tc<=ttt;tc++){
			long l = sc.nextLong();
			long r = sc.nextLong();
			if(l == r) {
				out.write(0+"\n");
				continue;
			}
			long find = 0;
			long x = r, y = l;
			while(l<=r) {
				long mid = (l+r)/2;
				if(2L*mid-1L<=x) {
//					System.out.println("hii");
					find = mid;
					l = mid+1;
				}
				else {
					r = mid-1;
				}
			}
			if(find == 0) {
				find = x%y+1;
			}
			out.write((find-1)+"\n");
		}
		out.close();
	}
	static long pow(long a, long b){long ret = 1;while(b>0){if(b%2 == 0){a = (a*a)%mod;b /= 2;}else{ret = (ret*a)%mod;b--;}}return ret%mod;}
	static long gcd(long a,long b){if(b==0) return  a; return gcd(b,a%b); } 
	private static void sort(double[] a) {List<Double> k = new ArrayList<>();for(double val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
	private static void ini(List<Integer>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
	private static void init(List<int[]>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
	private static void sort(long[] a) {List<Long> k = new ArrayList<>();for(long val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
}