Each evening after the dinner the SIS's students gather together to play the game of Sport Mafia. 

For the tournament, Alya puts candies into the box, which will serve as a prize for a winner. To do that, she performs n actions. The first action performed is to put a single candy into the box. For each of the remaining moves she can choose from two options:

  * the first option, in case the box contains at least one candy, is to take exactly one candy out and eat it. This way the number of candies in the box decreased by 1; 
  * the second option is to put candies in the box. In this case, Alya will put 1 more candy, than she put in the previous time. 



Thus, if the box is empty, then it can only use the second option.

For example, one possible sequence of Alya's actions look as follows:

  * put one candy into the box; 
  * put two candies into the box; 
  * eat one candy from the box; 
  * eat one candy from the box; 
  * put three candies into the box; 
  * eat one candy from the box; 
  * put four candies into the box; 
  * eat one candy from the box; 
  * put five candies into the box; 



This way she will perform 9 actions, the number of candies at the end will be 11, while Alya will eat 4 candies in total.

You know the total number of actions n and the number of candies at the end k. You need to find the total number of sweets Alya ate. That is the number of moves of the first option. It's guaranteed, that for the given n and k the answer always exists.

Please note, that during an action of the first option, Alya takes out and eats exactly one candy.

Input

The first line contains two integers n and k (1 ≤ n ≤ 10^9; 0 ≤ k ≤ 10^9) — the total number of moves and the number of candies in the box at the end. 

It's guaranteed, that for the given n and k the answer exists.

Output

Print a single integer — the number of candies, which Alya ate. Please note, that in this problem there aren't multiple possible answers — the answer is unique for any input data. 

Examples

Input


1 1


Output


0

Input


9 11


Output


4

Input


5 0


Output


3

Input


3 2


Output


1

Note

In the first example, Alya has made one move only. According to the statement, the first move is always putting one candy in the box. Hence Alya ate 0 candies.

In the second example the possible sequence of Alya's actions looks as follows: 

  * put 1 candy, 
  * put 2 candies, 
  * eat a candy, 
  * eat a candy, 
  * put 3 candies, 
  * eat a candy, 
  * put 4 candies, 
  * eat a candy, 
  * put 5 candies. 



This way, she will make exactly n=9 actions and in the end the box will contain 1+2-1-1+3-1+4-1+5=11 candies. The answer is 4, since she ate 4 candies in total.
''' CODED WITH LOVE BY SATYAM KUMAR '''

from sys import stdin, stdout
import cProfile, math
from collections import Counter,defaultdict,deque
from bisect import bisect_left,bisect,bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
import operator as op
from functools import reduce
sys.setrecursionlimit(10**6) # max depth of recursion
threading.stack_size(2**27)  # new thread will get stack of such size
fac_warmup = False
printHeap = str()
memory_constrained = False
P = 10**9+7
import sys

class merge_find:
    def __init__(self,n):
        self.parent = list(range(n))
        self.size = [1]*n
        self.num_sets = n
        self.lista = [[_] for _ in range(n)]
    def find(self,a):
        to_update = []
        while a != self.parent[a]:
            to_update.append(a)
            a = self.parent[a]
        for b in to_update:
            self.parent[b] = a
        return self.parent[a]
    def merge(self,a,b):
        a = self.find(a)
        b = self.find(b)
        if a==b:
            return
        if self.size[a]<self.size[b]:
            a,b = b,a
        self.num_sets -= 1
        self.parent[b] = a
        self.size[a] += self.size[b]
        self.lista[a] += self.lista[b]
    def set_size(self, a):
        return self.size[self.find(a)]
    def __len__(self):
        return self.num_sets

def display(string_to_print):
    stdout.write(str(string_to_print) + "\n")

def primeFactors(n): #n**0.5 complex 
    factors = dict()
    for i in range(2,math.ceil(math.sqrt(n))+1):  
        while n % i== 0: 
            if i in factors:
                factors[i]+=1
            else: factors[i]=1
            n = n // i 
    if n>2:
        factors[n]=1
    return (factors)

def all_factors(n):    
    return set(reduce(list.__add__, 
                ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))

def fibonacci_modP(n,MOD):
    if n<2: return 1
    #print (n,MOD)
    return (cached_fn(fibonacci_modP, (n+1)//2, MOD)*cached_fn(fibonacci_modP, n//2, MOD) + cached_fn(fibonacci_modP, (n-1) // 2, MOD)*cached_fn(fibonacci_modP, (n-2) // 2, MOD)) % MOD

def factorial_modP_Wilson(n , p): 
    if (p <= n): 
        return 0
    res = (p - 1) 
    for i in range (n + 1, p): 
        res = (res * cached_fn(InverseEuler,i, p)) % p 
    return res 

def binary(n,digits = 20):
    b = bin(n)[2:]
    b = '0'*(digits-len(b))+b
    return b

def isprime(n):
    """Returns True if n is prime."""
    if n < 4:
        return True
    if n % 2 == 0:
        return False
    if n % 3 == 0:
        return False
    i = 5
    w = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i += w
        w = 6 - w
    return True

def generate_primes(n):
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * 2, n+1, p): 
                prime[i] = False
        p += 1
    return prime

factorial_modP = []
def warm_up_fac(MOD):
    global factorial_modP,fac_warmup
    if fac_warmup: return
    factorial_modP= [1 for _ in range(fac_warmup_size+1)]
    for i in range(2,fac_warmup_size):
        factorial_modP[i]= (factorial_modP[i-1]*i) % MOD
    fac_warmup = True

def InverseEuler(n,MOD):
    return pow(n,MOD-2,MOD)

def nCr(n, r, MOD):
    global fac_warmup,factorial_modP
    if not fac_warmup:
        warm_up_fac(MOD)
        fac_warmup = True
    return (factorial_modP[n]*((pow(factorial_modP[r], MOD-2, MOD) * pow(factorial_modP[n-r], MOD-2, MOD)) % MOD)) % MOD

def test_print(*args):
    if testingMode:
        print(args)

def display_list(list1, sep=" "):
    stdout.write(sep.join(map(str, list1)) + "\n")

def display_2D_list(li):
    for i in li:
        print(i)
def prefix_sum(li):
    sm = 0
    res = []
    for i in li:
        sm+=i
        res.append(sm)
    return res

def get_int():
    return int(stdin.readline().strip())

def get_tuple():
    return map(int, stdin.readline().split())

def get_list():
    return list(map(int, stdin.readline().split()))
import heapq,itertools
pq = []                         # list of entries arranged in a heap
entry_finder = {}               # mapping of tasks to entries
REMOVED = '<removed-task>' 
def add_task(task, priority=0):
    'Add a new task or update the priority of an existing task'
    if task in entry_finder:
        remove_task(task)
    count = next(counter)
    entry = [priority, count, task]
    entry_finder[task] = entry
    heapq.heappush(pq, entry)

def remove_task(task):
    'Mark an existing task as REMOVED.  Raise KeyError if not found.'
    entry = entry_finder.pop(task)
    entry[-1] = REMOVED

def pop_task():
    'Remove and return the lowest priority task. Raise KeyError if empty.'
    while pq:
        priority, count, task = heapq.heappop(pq)
        if task is not REMOVED:
            del entry_finder[task]
            return task
    raise KeyError('pop from an empty priority queue')
memory = dict()
def clear_cache():
    global memory
    memory = dict()
def cached_fn(fn, *args):
    global memory
    if args in memory:
        return memory[args]
    else:
        result = fn(*args)
        memory[args] = result
        return result

def ncr (n,r):
    return math.factorial(n)/(math.factorial(n-r)*math.factorial(r))
def binary_serach(i,li):
    #print("Search for ",i)
    fn = lambda x: li[x]-x//i
    x = -1
    b = len(li)
    while b>=1:
        #print(b,x)
        while b+x<len(li) and fn(b+x)>0: #Change this condition 2 to whatever you like
            x+=b
        b=b//2
    return x

# -------------------------------------------------------------- MAIN PROGRAM
TestCases = False
testingMode = False
fac_warmup_size = 10**5+100
optimiseForReccursion = False #Can not be used clubbed with TestCases # WHen using recursive functions, use Python 3
from math import factorial

def main():
    n, k = get_tuple()
    x = 0
    while (x*(x+1))//2<k:
        x+=1
    ate = ((x*(x+1))//2)-k
    n -= x+ate
    while n>0:
        x+=1
        ate+=x
        n-=(x+1)
    print(ate)

# --------------------------------------------------------------------- END=


if TestCases: 
    for i in range(get_int()): 
        cProfile.run('main()') if testingMode else main() 
else: (cProfile.run('main()') if testingMode else main()) if not optimiseForReccursion else threading.Thread(target=main).start()