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:
#define _USE_MATH_DEFINES
#include<iostream>
#include<string>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<list>
#include<iomanip>
#include<vector>
#include<random>
#include<functional>
#include<algorithm>
#include<cstdio>
#include<unordered_map>
using namespace std;
//---------------------------------------------------
//ライブラリゾーン！！！！
typedef long long ll;
typedef long double ld;
#define str string
#define rep(i,j) for(ll i=0;i<(long long)(j);i++)
const ll Mod = 1000000007;
const ll gosenchou = 5000000000000000;
short gh[2][4] = { { 0,0,-1,1 },{ -1,1,0,0 } };
struct P {
	ll pos, cost;
};
bool operator<(P a, P b) { return a.cost < b.cost; }
bool operator>(P a, P b) { return a.cost > b.cost; }
struct B {//隣接リスト表現
	ll to;
	ld cost;
};
struct E {//辺の情報を入れる変数
	ll from, to, cost;
};
bool operator<(E a, E b) {
	return a.cost < b.cost;
}
struct H {
	ll x, y;
};
bool operator<(H a, H b) {
	/*if (a.x != b.x) return a.x < b.x;
	return a.y < b.y;*/
	return ((a.x + 1)*(a.y + 1)) < ((b.x + 1)*(b.y + 1));
}
bool operator>(H a, H b) {
	if (a.x != b.x) return a.x > b.x;
	return a.y > b.y;
}
bool operator==(H a, H b) {
	return a.x == b.x&&a.y == b.y;
}
bool operator!=(H a, H b) {
	return a.x != b.x || a.y != b.y;
}
ll gcm(ll i, ll j) {//最大公約数
	if (i > j) swap(i, j);
	if (i == 0) return j;
	return gcm(j%i, i);
}
ld rad(H a, H b) {
	return sqrt(pow(a.x - b.x, 2.0) + pow(a.y - b.y, 2.0));
}//rad＝座標上の2点間の距離
ll ari(ll a, ll b, ll c) {
	return (a + b)*c / 2;
}//等差数列の和
bool suf(ld a, ld b, ld c, ld d) {
	if (b <= c || d <= a) return 0;
	return 1;
}//[a,b),[c,d)
ll fact(ll x, ll k, ll p) {//最大値、個数
	ll sum = 1;
	for (int i = 0; i < k; i++) {
		sum *= (x--);
		sum %= p;
	}
	return sum;
}//階乗(正）
ll mod_pow(ll x, ll n, ll p) {
	ll res = 1;
	while (n > 0) {
		if (n & 1) res = res*x%p;
		x = x*x%p;
		n >>= 1;
	}
	return res;
}//x^n%p
 //#define int long long
const long long inf = 4523372036854775807;
const int iinf = 1500000000;
//----------------------------------------------------
//++++++++++++++++++++++++++++++++++++++++++++++++++++
ld d, ans;
ld MAX(ld x) {
	ld y = sqrt(d*d - x*x);
	return x + y;
}
signed main() {
	cin >> d;
	ans = d;
	for (int x = 1; x <= 20; x++) {
		ld y = d*d - x*x;
		if (y < 0) break;
		if (y <= 1) ans = max(ans, (ld)x + 1);
		else ans = max(ans, x + sqrt(y));
	}
	ld s = 0, e = d, t1, t2;
	while (e - s > 0.0000000000001) {
		t1 = s + (e - s) / 3, t2 = s + (e - s) / 3 * 2;
		ld r1 = MAX(t1), r2 = MAX(t2);
		if (r1 == r2) s = t1, e = t2;
		else if (r1 > r2) e = t2;
		else s = t1;
	}
	printf("%.15LF\n", max(MAX(s),ans));
	getchar(); getchar();
}