CF351A Jeff and Rounding 思维

Jeff got 2n real numbers a₁, a₂, ..., a_2n as a birthday present. The boy hates non-integer numbers, so he decided to slightly "adjust" the numbers he's got. Namely, Jeff consecutively executes n operations, each of them goes as follows:

• choose indexes i and j (i ≠ j) that haven't been chosen yet;
• round element a_i to the nearest integer that isn't more than a_i (assign to a_i: ⌊ a_i ⌋);
• round element a_j to the nearest integer that isn't less than a_j (assign to a_j: ⌈ a_j ⌉).

Nevertheless, Jeff doesn't want to hurt the feelings of the person who gave him the sequence. That's why the boy wants to perform the operations so as to make the absolute value of the difference between the sum of elements before performing the operations and the sum of elements after performing the operations as small as possible. Help Jeff find the minimum absolute value of the difference.

Input

The first line contains integer n (1 ≤ n ≤ 2000). The next line contains 2n real numbers a₁, a₂, ..., a_2n (0 ≤ a_i ≤ 10000), given with exactly three digits after the decimal point. The numbers are separated by spaces.

Output

In a single line print a single real number — the required difference with exactly three digits after the decimal point.

Examples

Input

Copy

3
0.000 0.500 0.750 1.000 2.000 3.000

Output

Copy

0.250

Input

Copy

3
4469.000 6526.000 4864.000 9356.383 7490.000 995.896

Output

Copy

0.279

Note

In the first test case you need to perform the operations as follows: (i = 1, j = 4), (i = 2, j = 3), (i = 5, j = 6). In this case, the difference will equal |(0 + 0.5 + 0.75 + 1 + 2 + 3) - (0 + 0 + 1 + 1 + 2 + 3)| = 0.25.

假设小数部分是x的话，向下取整为-x,向上为1-x;

可以发现不论是向下还是向上都是 -x，那么小数部分就可以统一处理；

那么问题就是当向上取整时，会+1----->求1的个数；

那么枚举就行了；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int n;
int m;

int main() {
	//ios::sync_with_stdio(0);
	cin >> n;
	m = 2 * n;
	double tmp;
	int numeq = 0;
	double sum = 0.0;
	int numdb = 0;
	for (int i = 1; i <= m; i++) {
		rdlf(tmp);
		ll intmp = (ll)tmp;
		if (intmp == tmp)numeq++;
		else {
			sum += 1.0*(tmp - intmp);
			numdb++;
		}
	}
	int minn = min(n, numeq);
	double ans = inf;
	for (int i = 0; i <= minn; i++) {
		ans = min(ans, (double)fabs(n - i - sum));
	}
	printf("%.3lf\n", 1.0*ans);
	return 0;
}