/*
với số Nguyên N ta phân tích:
nếu muốn cho số đó 
trừng hợp 1: N là số chính phương. ta xuất ra 0 vì sqrt(n)-sqrt(n)=0.
trùng hợp còn lại: ta chạy trừng từ căn bậc 2 của N nếu i*n/i=n thi ta xuat ra i-n/.
*/
#pragma GCC optimize("O3","unroll-loops")
#include <bits/stdc++.h>
#define ll long long
#define endl "\n"
#define lcm(a, b) (a * b) / __gcd(a, b);
#define sq(n) (ll)sqrt(n)
using namespace std;
const ll INF = 1e18;
const ll mod = 1e9 + 7;
int main() {
  ios :: sync_with_stdio(false);
  cin.tie(nullptr);
  ll n;
  cin >> n;
  n = abs(n);
  if(sq(n) * sq(n) == n) {
     cout << 0;
     return 0;
  }
  for(ll i = sq(n); i >= 1; i --) {
    if(i * (n / i) == n) {
      cout << abs(i - (n / i));
      return 0;
    }
  }
}
 

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