      
    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;
long long conc(long long a, long long b) {
  string s1;
  string s2;
  if (a == 0)
    s1 = "0";
  else
    while (a) {
      s1 = (char(a % 10 + '0')) + s1;
      a /= 10;
    }
  if (b == 0)
    s2 = "0";
  else
    while (b) {
      s2 = (char(b % 10 + '0')) + s2;
    }
  s1 += s2;
  long long c = 0, d = 1;
  for (int i = s1.length() - 1; i >= 0; i--) {
    c += (s1[i] - '0') * d;
    d *= 10;
  }
  return c;
}
int main() {
  int x;
  cin >> x;
  double foo = 0;
  double bar = 0;
  double baz = 0;
  double quz = 1;
  double pur;
  while (x > 0) {
    cin >> pur;
    foo = foo + pur;
    bar = bar + 1;
    x--;
    if (max(foo * quz, bar * baz) == foo * quz) {
      baz = foo;
      quz = bar;
    }
  }
  printf("%.8f", baz / quz);
}
