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;
void _fill_int(int* p, int val, int rep) {
  int i;
  for (i = 0; i < rep; i++) p[i] = val;
}
int GETi() {
  int i;
  scanf("%d", &i);
  return i;
}
template <class T>
T sqr(T val) {
  return val * val;
}
void solve() {
  int f, r, i, j, k, l, x, y;
  int ma = 0;
  int TUX, FOO, BAR, BAZ, QUZ, PUR;
  TUX = GETi();
  FOO = BAR = BAZ = 0;
  QUZ = 1;
  while (TUX--) {
    PUR = GETi();
    FOO += PUR;
    BAR++;
    if (FOO * QUZ >= BAR * BAZ) {
      BAZ = FOO;
      QUZ = BAR;
    }
  }
  printf("%.12lf\n", BAZ / (double)QUZ);
  return;
}
int main(int argc, char** argv) {
  if (argc > 1) freopen(argv[1], "r", stdin);
  solve();
  return 0;
}