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.*; // Implicit import
import java.io.*;
import java.math.BigInteger; // Explicit import

public class A {

	static FastReader		sc	= new FastReader();
	static PrintWriter	out	= new PrintWriter(System.out);


	public static void main(String[] args) throws Exception {

		int t = sc.ni();
		while (t-- > 0) {
			A.go();
		}
		out.flush();

	}

	// >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Code Starts <<<<<<<<<<<<<<<<<<<<<<<<<<<<< //
 static void go() {
	 long l=sc.ni();
	 long r=sc.ni();
	 if(l==r) {
		 out.println(0);
		 return;
	 }
	// a=b+x;
	if((r-l)==1) {
		out.println(r%l);
		return;
	}
	long rr=r/2;

	rr++;
	out.println((r)%rr);
	
}

	// >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Code Ends <<<<<<<<<<<<<<<<<<<<<<<<<<<<< //
	static long	fact[];
	static long	invfact[];

	static long ncr(int n, int k) {
		if (k < 0 || k > n) {
			return 0;
		}
		long x = fact[n] % mod;
		long y = invfact[k] % mod;
		long yy = invfact[n - k] % mod;
		long ans = (x * y) % mod;
		ans = (ans * yy) % mod;
		return ans;
	}

	static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	static int	prime[]	= new int[200005];
	static int	N				= 200005;

	static void sieve() {

		for (int i = 0; i < N; i++) {
			prime[i] = i;
		}
		for (int i = 2; i * i <= N; i++) {
			if (prime[i] == i) {
				prime[i] = i;
				for (int j = i; j < N; j += i) {
					prime[j] = i;
				}
			}
		}
	}

	static int gcd(int a, int b) {
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	static void sort(int[] a) {
		ArrayList<Integer> aa = new ArrayList<>();
		for (Integer i : a) {
			aa.add(i);
		}
		Collections.sort(aa);
		for (int i = 0; i < a.length; i++)
			a[i] = aa.get(i);
	}

	static void sort(long[] a) {
		ArrayList<Long> aa = new ArrayList<>();
		for (Long i : a) {
			aa.add(i);
		}
		Collections.sort(aa);
		for (int i = 0; i < a.length; i++)
			a[i] = aa.get(i);
	}

	static int mod = (int) 1000000007;

	static long pow(long x, long y) {
		long res = 1l;
		while (y != 0) {
			if (y % 2 == 1) {
				res = (x % mod * res % mod) % mod;
			}
			y /= 2;
			x = (x % mod * x % mod) % mod;
		}
		return res % mod;
	}

	// >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Fast IO <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< //

	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 ni() {
			return Integer.parseInt(next());
		}

		long nl() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		int[] intArray(int n) {
			int a[] = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = sc.ni();
			return a;
		}

		long[] longArray(int n) {
			long a[] = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = sc.nl();
			return a;
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}