Problem:
Score : 200 points

Problem Statement

There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0.
During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right.
That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i.
The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible.
Find the earliest possible time to reach coordinate X.

Constraints

X is an integer.

1≤X≤10^9

Input

The input is given from Standard Input in the following format:

X

Output

Print the earliest possible time for the kangaroo to reach coordinate X.

Sample Input 1

6

Sample Output 1

3

The kangaroo can reach his nest at time 3 by jumping to the right three times, which is the earliest possible time.

Sample Input 2

2

Sample Output 2

2

He can reach his nest at time 2 by staying at his position during the first second, and jumping to the right at the next second.

Sample Input 3

11

Sample Output 3

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES

#include <bits/stdc++.h>
using namespace std;
using i32 = int; using i64 = long long int; using f64 = double; using str = string;
template <typename T> using vec = vector<T>;
template <typename T> using heap = priority_queue<T, vec<T>, greater<T>>;
#define times(n, i) for (i32 i = 0; i < (n); ++i)
#define range(a, b, i) for (i32 i = (a); i < (b); ++i)
#define upto(a, b, i) for (i32 i = (a); i <= (b); ++i)
#define downto(a, b, i) for (i32 i = (a); i >= (b); --i)
#define all(xs) (xs).begin(), (xs).end()
#define sortall(xs) sort(all(xs))
#define reverseall(xs) reverse(all(xs))
#define uniqueall(xs) (xs).erase(unique(all(xs)), (xs).end())
#define even(x) (((x) & 1) == 0)
#define odd(x) (((x) & 1) == 1)
#define append emplace_back
const i64 MOD = 1000000007;

i32 x;

i32 main()
{
    cin >> x;
    i32 acc = 0;
    upto(1, x, i) {
        acc += i;
        if (acc >= x) {
            cout << i << endl;
            break;
        }
    }
    return 0;
}