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
import math
import itertools
import functools
import collections
import operator
import fileinput
import copy
import string

ORDA = 97  # a


def ii(): return int(input())


def mi(): return map(int, input().split())


def li(): return list(map(int, input().split()))


def lcm(a, b): return abs(a * b) // math.gcd(a, b)


def revn(n): return str(n)[::-1]


def dd(): return collections.defaultdict(int)


def ddl(): return collections.defaultdict(list)


def sieve(n):
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r


def divs(n, start=2):
    r = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if (n % i == 0):
            if (n // i == i):
                r.append(i)
            else:
                r.extend([i, n // i])
    return r


def divn(n, primes):
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t


def prime(n):
    if n == 2: return True
    if n % 2 == 0 or n <= 1: return False
    sqr = int(math.sqrt(n)) + 1
    for d in range(3, sqr, 2):
        if n % d == 0: return False
    return True


def convn(number, base):
    new_number = 0
    while number > 0:
        new_number += number % base
        number //= base
    return new_number


def cdiv(n, k): return n // k + (n % k != 0)


def ispal(s):
    for i in range(len(s) // 2 + 1):
        if s[i] != s[-i - 1]:
            return False
    return True


for _ in range(ii()):
    l, r = mi()
    if 2 * l > r:
        print(r % l)
    else:
        print(r // 2 if r & 1 else r // 2 - 1)

