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;
template <typename T>
void read(T &x) {
  T f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  x *= f;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
double tux, pur, foo = 0, bar = 0, baz = 0, quz = 1;
int main() {
  scanf("%lf", &tux);
  for (int i = 1; i <= tux; ++i) {
    scanf("%lf", &pur);
    foo += pur;
    bar++;
    if (foo * quz > baz * bar) {
      baz = foo;
      quz = bar;
    }
  }
  printf("%.6lf", baz / quz);
  return 0;
}