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 bisect
import math

mod =  998244353
N = 200005

def mul(a, b):
	return (a*b)%mod

def add(a, b):
	return (a+b) if (a+b<mod) else (a+b)-mod

def sub(a, b):
	return (a-b) if (a-b>0) else (a-b)+mod

def powr(a, b):
	ans = 1
	while b>0:
		if b & 1:
			ans=mul(ans,a)
		a = mul(a,a) 
		b//=2
	return ans

def inv(n):
	return powr(n, mod-2)

def factlist():
	fact = [1]
	for i in range(1, N):
		fact.append(mul(fact[i-1],i))
	return fact

def invfactlist(fact):
	invfact=[0]*N
	invfact[0]=1
	invfact[N-1]= inv(fact[N-1])
	for i in range(N-2, 0, -1):
		invfact[i]= mul(invfact[i+1],(i+1))
	
	return invfact

def rot(S):
	return list(zip(*S[::-1]))

def gcd(a, b):
	if b==0:
		return a
	return gcd(b, a%b)


def main():
	t = int(input())
	while t:
		t-=1 

		l, r = map(int, input().split())
		if l == r:
			print('0')
			continue
		if l>(r+1)//2:
			print(r-l)
			continue
		else:
			print(r-(r+1)//2-1)

	

if __name__ == "__main__":
	main()
	
