      
    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;
string t[] = {"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"};
int main() {
  int tux;
  cin >> tux;
  int foo = 0, bar = 0, baz = 0, quz = 1;
  int pur;
  for (; tux != 0; tux--) {
    cin >> pur;
    foo += pur;
    bar += 1;
    if (max((foo * quz), (bar * baz)) == (foo * quz)) {
      baz = foo;
      quz = bar;
    }
  }
  cout << std::setprecision(9) << baz / (double)quz << "\n";
  return 0;
}
