      
    HAI  
    I HAS A TUX  
    GIMMEH TUX  
    I HAS A FOO ITS 0  
    I HAS A BAR ITS 0  
    I HAS A BAZ ITS 0  
    I HAS A QUZ ITS 1  
    TUX IS NOW A NUMBR  
    IM IN YR LOOP NERFIN YR TUX TIL BOTH SAEM TUX AN 0  
    I HAS A PUR  
    GIMMEH PUR  
    PUR IS NOW A NUMBR  
    FOO R SUM OF FOO AN PUR  
    BAR R SUM OF BAR AN 1  
    BOTH SAEM BIGGR OF PRODUKT OF FOO AN QUZ AN PRODUKT OF BAR BAZ AN PRODUKT OF FOO AN QUZ  
    O RLY?  
    YA RLY  
    BAZ R FOO  
    QUZ R BAR  
    OIC  
    IM OUTTA YR LOOP  
    BAZ IS NOW A NUMBAR  
    VISIBLE SMOOSH QUOSHUNT OF BAZ QUZ  
    KTHXBYE  
    

Input

The input contains between 1 and 10 lines, i-th line contains an integer number xi (0 ≤ xi ≤ 9).

Output

Output a single real number. The answer is considered to be correct if its absolute or relative error does not exceed 10 - 4.

Examples

Input

3
0
1
1


Output

0.666667
#include <bits/stdc++.h>
using namespace std;
int ToInt(string& s, int len) {
  int r = 0;
  for (int i = 0, max_ = (len); i < max_; ++i) {
    r *= 10;
    r += s[i] - '0';
  }
  return r;
}
int GCD(int a, int b) { return b != 0 ? GCD(b, a % b) : a; }
int LCM(int a, int b) { return a * (b / GCD(a, b)); }
long long Pow(long long n, long long e) {
  if (e == 0) return 1;
  if (e == 1)
    return n;
  else if (e & 1) {
    long long t = Pow(n, e / 2);
    return n * t * t;
  } else {
    long long t = Pow(n, e / 2);
    return t * t;
  }
}
int main() {
  double max = 0.0;
  int N, sum = 0, t;
  scanf("%d", &N);
  for (int i = 0, max_ = (N); i < max_; ++i) {
    scanf("%d", &t);
    sum += t;
    max = std::max(max, sum / (double)(i + 1));
  }
  printf("%.8lf\n", max);
  return 0;
}
