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.
# cook your dish here
#import sys
#sys.setrecursionlimit(10**9) 
ll=lambda:map(int,input().split())
t=lambda:int(input())
ss=lambda:input()
lx=lambda x:map(int,input().split(x))
yy=lambda:print("YES")
nn=lambda:print("NO")

from math import log10 ,log2,ceil,factorial as fac,gcd,inf,sqrt,log
#from itertools import combinations_with_replacement as cs 
#from functools import reduce
from bisect import bisect_right as br,bisect_left as bl
from collections import Counter
#from math import inf


mod=10**9+7



     

for _ in range(t()):
    
#def f():
    
    
    a,b=ll()

    x=b//2+(1 if b%2==0 else 0)
    if 2*a>b:
        print(b-a)
    else:
        print((b-1)//2)
                
#f()
    

'''
3 2 1 1 1 3 2 1
baca

bac
1 2 3
baaccca

abbaccccaba

1 1111
2 11
3 11
 
'''

