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++ (GCC 9.2.1) (judge verdict: Accepted)

// 　　＿＿
// 　／　<＠ﾌ
//  ｜(ﾉﾉハ))
//  ﾉ从ﾟヮﾟ从
// 　ﾉ｜ｿﾉГ|つ author:hotarunx
// 〈_ﾉ^^^ヽ|
// 　~~tｧtｧ~
#include <algorithm>
#include <array>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <vector>
using namespace std;
#define int long long

// 時刻iにカンガルーは座標0から座標1+2+...+iにたどり着くことができる。
// 時刻iを小さい順に試して可能か調べる

signed main() {
    cin.tie(0);
    ios::sync_with_stdio(0);

    int x;
    cin >> x;

    int cumsumx = 0;

    for (int i = 1;; i++) {
        cumsumx += i;
        if (cumsumx >= x) {
            cout << i << endl;
            return 0;
        }
    }
}