#!/usr/bin/env python3


""" Using high school algebra to invert

        k \mapsto k * (k - 1)/2,

    we arrive at

        h = \lfloor \frac{ -1 \pm \sqrt{ 1 + 8 k }{2} \rfloor
"""

import math


def invert_formula(n) -> float:
    # return the positive root
    a = 1
    b = 1
    c = -2 * n
    discriminant = b**2 - 4 * a * c
    assert discriminant >= 0
    root1 = (-b + math.sqrt(discriminant)) / (2 * a)
    root2 = (-b - math.sqrt(discriminant)) / (2 * a)
    return max(root1, root2)


print(math.floor(invert_formula(int(input()))))