Problem: In Manhattan, roads run where the x or y coordinate is an integer. Both Sunuke-kun's house and Sumeke-kun's house are on the road, and the straight line distance (Euclidean distance) is just d. Find the maximum value that can be considered as the shortest distance when traveling along the road from Sunuke-kun's house to Sumek-kun's house.

Constraints

* 0 <d ≤ 10
* d is given up to exactly three decimal places

Input


d


Output

Print the answer on one line. If the absolute error or relative error is 10-9 or less, the answer is judged to be correct.

Examples

Input

1.000


Output

2.000000000000


Input

2.345


Output

3.316330803765

Solution (in CPP):

Solution:
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#include <climits>
#include <cstring>
#include <cassert>

#define rep(i, m, n) for(int i=int(m);i<int(n);i++)
#define all(c) begin(c),end(c)

template<typename T1, typename T2>
inline void chmin(T1 &a, T2 b) { if (a > b) a = b; }

template<typename T1, typename T2>
inline void chmax(T1 &a, T2 b) { if (a < b) a = b; }

typedef long long int ll;
using ll = long long int;
using ull = long long unsigned int;
using Int = long long int;
using uInt = long long unsigned int;
using Double = long double;
using namespace std;
#define INF (1 << 30) - 1
#define INFl (ll)5e15
#define DEBUG 0
#define dump(x)  cerr << #x << " = " << (x) << endl
#define MOD 1000000007


//edit
class Solve {
public:
    double d;

    Double calc(Double x) {
        if (d * d - x * x < 0) return 0.0;

        Double y = sqrt(d * d - x * x);
        Double ret = x + max(1.l, y);
        return ret;
    }

    void solve() {
        cin >> d;
        Double ans = d;
        chmax(ans, sqrt(2.) * d);
        for (int i = 0; i <= 10; ++i) {
//            cerr << calc(i) << endl;
            chmax(ans, calc(i));
        }
        cout << ans << endl;
    }
};


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
//    std::ifstream in("input.txt");
//    std::cin.rdbuf(in.rdbuf());

    Solve().solve();


    return 0;
}