#include<iostream>
using namespace std;
const int N = 1e6 + 10;
double a[N];
int main() {
	double max1, min1 = 1e6;
	double sum = 0, aver = 0;
	long long n;
	cin >> n ;
	for (int i =0 ; i < n; i++) {
		cin >> a[i];
		max1 = max(max1, a[i]);
		min1 = min(min1, a[i]);
		sum += a[i];
		if (i >= 2) {
			aver = (sum - max1 - min1) / (i - 1);
			printf("%.2lf\n", aver);
		}
	}
	return 0;
}