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.
"""
Author    : co_devil Chirag Garg
Institute : JIIT
"""


from __future__ import division, print_function
from sys import stdin,stdout
import itertools, os, sys, threading
from collections import deque, Counter, OrderedDict, defaultdict
import heapq
from math import ceil,floor,log,sqrt,factorial,pow,pi,gcd
# from bisect import bisect_left,bisect_right
# from decimal import *,threading
"""from io import BytesIO, IOBase
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
else:
    from builtins import str as __str__
    str = lambda x=b'': x if type(x) is bytes else __str__(x).encode()
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._buffer = BytesIO()
        self._fd = file.fileno()
        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):
        return self._buffer.read() if self._buffer.tell() else os.read(self._fd, os.fstat(self._fd).st_size)

    def readline(self):
        while self.newlines == 0:
            b, ptr = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)), self._buffer.tell()
            self._buffer.seek(0, 2), self._buffer.write(b), self._buffer.seek(ptr)
            self.newlines += b.count(b'\n') + (not b)
        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)
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip(b'\r\n')

def print(*args, **kwargs):
    sep, file = kwargs.pop('sep', b' '), kwargs.pop('file', sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop('end', b'\n'))
    if kwargs.pop('flush', False):
        file.flush()

"""


def ii(): return int(input())
def si(): return str(input())
def mi(): return map(int,input().split())
def li(): return list(mi())

def fii(): return int(stdin.readline())
def fsi(): return str(stdin.readline())
def fmi(): return map(int,stdin.readline().split())
def fli(): return list(fmi())


abc = 'abcdefghijklmnopqrstuvwxyz'
abd = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12,
       'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24,
       'z': 25}
mod = 1000000007
dx, dy = [-1, 1, 0, 0], [0, 0, 1, -1]
def getKey(item): return item[0]
def sort2(l): return sorted(l, key=getKey)
def d2(n, m, num): return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo(x): return (x and (not (x & (x - 1))))
def decimalToBinary(n): return bin(n).replace("0b", "")
def ntl(n): return [int(i) for i in str(n)]
def powerMod(x, y, p):
    res = 1
    x %= p
    while y > 0:
        if y & 1:
            res = (res * x) % p
        y = y >> 1
        x = (x * x) % p
    return res
def gcd(x, y):
    while y:
        x, y = y, x % y
    return x
# For getting input from input.txt file
# sys.stdin = open('input.txt', 'r')

# Printing the Output to output.txt file
# sys.stdout = open('output.txt', 'w')

graph = defaultdict(list)
visited = [0] * 1000000
col = [-1] * 1000000
def dfs(v, c):
    if visited[v]:
        if col[v] != c:
            print('-1')
            exit()
        return
    col[v] = c
    visited[v] = 1
    for i in graph[v]:
        dfs(i, c ^ 1)

def bfs(d,v):
    q=[]
    q.append(v)
    visited[v]=1
    while len(q)!=0:
        x=q[0]
        q.pop(0)
        for i in d[x]:
            if visited[i]!=1:
                visited[i]=1
                q.append(i)
        print(x)
    print(l)

def make_graph(e):
    d={}
    for i in range(e):
        x,y=mi()
        if x not in d.keys():
            d[x]=[y]
        else:
            d[x].append(y)
        if y not in d.keys():
            d[y] = [x]
        else:
            d[y].append(x)
    return d
def gr2(n):
    d={}
    for i in range(n):
        x,y=mi()
        if x not in d.keys():
            d[x]=[y]
        else:
            d[x].append(y)
    return d

def connected_components(graph):
    seen = set()
    def dfs(v):
        vs = set([v])
        component=[]
        while vs:
            v = vs.pop()
            seen.add(v)
            vs |= set(graph[v]) - seen
            component.append(v)
        return component
    ans=[]
    for v in graph:
        if v not in seen:
            d=dfs(v)
            ans.append(d)
    return ans

def primeFactors(n):
    s=set()
    while n % 2 == 0:
        s.add(2)
        n = n // 2
    for i in range(3, int(sqrt(n)) + 1, 2):
        while n % i == 0:
            s.add(i)
            n = n // i
    if n > 2:
        s.add(n)
    return s

n,k=mi()
c=0
p=0
s=0
while s-n<k:
    p+=1
    s+=p
    n-=1
print(n)