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 os
import sys
from io import BytesIO, IOBase
#import sys
#input = sys.stdin.buffer.readline
BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#######################################
from itertools import permutations
from collections import *
from collections import deque
from operator import itemgetter , attrgetter
from decimal import *
import bisect
import math
import heapq as hq
#import sympy
MOD=10**9 +7


def is_prime(n):
  if n == 2 or n == 3: return True
  if n < 2 or n%2 == 0: return False
  if n < 9: return True
  if n%3 == 0: return False
  r = int(n**0.5)
  # since all primes > 3 are of the form 6n ± 1
  # start with f=5 (which is prime)
  # and test f, f+2 for being prime
  # then loop by 6.
  f = 5
  while f <= r:

    if n % f == 0: return False
    if n % (f+2) == 0: return False
    f += 6
  return True

def pow(a,b,m):
    ans=1
    while b:
        if b&1:
            ans=(ans*a)%m
        b//=2
        a=(a*a)%m
    return ans
#vis=[]
#graph=[]

def ispalindrome(s):
    if s[:]==s[::-1]:
        return 1
    return 0
dp=[]
limit=[]
v=[]
def dpdfs(u,t=-1):
    dp[0][u]=0
    dp[1][u]=0
    for i in v[u]:
        if i==t:

            continue
        if dp[1][i]==-1:
            dpdfs(i,u)
        dp[0][u]+=max(abs(limit[0][u]-limit[1][i])+dp[1][i],abs(limit[0][u]-limit[0][i])+dp[0][i])
        dp[1][u] += max(abs(limit[1][u] - limit[1][i]) + dp[1][i], abs(limit[1][u] - limit[0][i]) + dp[0][i])
vis=[]
f=0
def dfs(i):
    vis[i]=1
    act[i]=1
    for j in v[i]:
        if act[j]:
            f=1
            #print(-1)
            return -1
        if vis[j]==0:
           if dfs(j)==-1:
               return -1
    act[i]=0
    ans.append(i)
    return 0
from queue import PriorityQueue

def z_algorithm(s):
    res = [0] * len(s)
    res[0] = len(s)
    i, j = 1, 0
    while i < len(s):
        while i + j < len(s) and s[j] == s[i + j]:
            j += 1
        res[i] = j
        if j == 0:
            i += 1
            continue
        k = 1
        while i + k < len(s) and k + res[k] < j:
            res[i + k] = res[k]
            k += 1
        i, j = i + k, j - k
    return res
def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)


# Function to return LCM of two numbers
def lcm(a, b):
    return (a / gcd(a, b)) * b


def gcdExtended(a, b):
    # Base Case
    if a == 0:
        return b, 0, 1

    gcd, x1, y1 = gcdExtended(b % a, a)

    # Update x and y using results of recursive
    # call
    x = y1 - (b // a) * x1
    y = x1

    return gcd, x, y
import re
for _ in range(int(input())):
    l,r=map(int,input().split())
    t=1+(r//2)
    if t>=l:
        print(r%t)
    else:
        print(r%l)






















