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.
from collections import deque, Counter
from collections import defaultdict as dfd
from bisect import bisect, bisect_left
from math import sqrt, gcd, ceil, factorial
from heapq import heapify, heappush, heappop

import sys
sys.setrecursionlimit(10**7) 

MOD = 10**9 + 7
inf = float("inf")
ans_ = []

def nin():return int(input())
def ninf():return int(file.readline())

def st():return (input().strip())
def stf():return (file.readline().strip())

def read(): return list(map(int, input().strip().split()))
def readf():return list(map(int, file.readline().strip().split()))

def readfl(): return list(map(float, input().strip().split()))
def readffl():return list(map(float, file.readline().strip().split()))





# file = open("input.txt", "r")
def solve():
    for _ in range(nin()):
        l, r = read()
        x = max(l, r//2 + 1)
        ans_.append(r%x)
    
solve()

for i in ans_:
    print(i)
    