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 sys,os
#if not os.environ.get('ONLINE_JUDGE'):
 #   sys.stdin = open('C:\\books and lectures\\cp-nptel\\in.txt', 'r')
  #  sys.stdout = open('C:\\books and lectures\\cp-nptel\\myout.txt', 'w')
import time
#start_time=time.time()
t=int(input())
while(t>0):
    l,r=list(map(int,input().split()))
    if(l==r):
        print(0)
    elif(l==r-1):
        print(1)
    elif(l==r//2):
        print(l-1)
    elif(l>r//2):
        print(r-l)
    else:
        if(r%2==0):
            print(r//2-1)
        else:
            print(r//2)
    t-=1
    
#print("---%s seconds ---"%{time.time()-start_time})
