#include <iostream>

using namespace std;
const int N = 1e5+2;
long long sum [N] ;
int main()
{
    int n ;
    long long ans;
    long long *srch;
    for (int i = 0 ; i<N ; ++i)
    {
        sum[i] = (1ll*i*(i+1))/2.0;
    }
    i:
    cin >> n ;
    n = (n<0 ? -n : n);
    srch = lower_bound(sum,sum+N,n);
    ans = srch - sum ;
    if (*srch != n) ++ans;
    if (*srch != n && n&1 ) ++ans;
    cout << ans <<endl ;
    goto i ;
    return 0;
}