Problem:       
    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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int tux;
double foo, bar, baz, quz = 1.0;
inline int rint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
int main() {
  tux = rint();
  for (register int i = (1), iend = (tux); i <= iend; i++) {
    double pur;
    scanf("%lf", &pur);
    foo = foo + pur;
    bar = bar + 1.0;
    if (foo * quz > baz * bar) {
      baz = foo;
      quz = bar;
    }
  }
  printf("%.6lf", baz / quz);
  return 0;
}