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)

#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
#define endl "\n"
#define sz(x) ((ll)(x).size())
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define prt(x) cout << (x)
#define prtl(x) cout << (x) << endl
#define rep(i,a,b) for(ll i=a; i<b; i++)
#define rrep(i,a,b) for(ll i=a; i>b; i--)
#define mp(a, b) make_pair(a, b)
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
using namespace std;
using vi = vector<int>;
using vll = vector<long long int>;
using vb = vector<bool>;
using vvb = vector<vector<bool>>;
using vc = vector<char>;
using vvi = vector<vector<int>>;
using vvll = vector<vector<long long int>>;
using vvc = vector<vector<char>>;
using ll = long long int;
using P = pair<long long int, long long int>;
using Map = map<long long int, long long int>;
ll INF = 1LL<<60;
ll M = 1000000007;

int main(){
    ll x; cin >> x;
    ll t = 0;
    ll n = 1;
    while(t < x){
        t += n;
        n++;
    }
    prt(n-1);
}