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;
long long convertToNum(string s) {
  long long val = 0;
  for (int i = 0; i < (s.size()); i++) val = val * 10 + s[i] - '0';
  return val;
}
char bu[50];
string convertToString(int a) {
  sprintf(bu, "%d", a);
  return string(bu);
}
long long GCD(long long x, long long y) {
  if (!x) return y;
  if (!y) return x;
  if (x == y) return x;
  if (x < y)
    return GCD(x, y % x);
  else
    return GCD(x % y, y);
}
long long POW(long long x, long long y, long long Base) {
  if (!y) return 1;
  long long u = POW(x, y / 2, Base);
  u = (u * u) % Base;
  if (y & 1)
    return (u * x) % Base;
  else
    return u;
}
void extended_euclid(long long A, long long B, long long &x, long long &y) {
  if (A == 1 && B == 0) {
    x = 1;
    y = 0;
    return;
  }
  if (A < B)
    extended_euclid(B, A, y, x);
  else {
    long long xx, yy;
    extended_euclid(A % B, B, xx, yy);
    x = xx;
    y = yy - (A / B) * xx;
  }
}
int n;
double PUR, FOO, BAR, BAZ, QUZ;
int main() {
  cin >> n;
  double FOO = 0;
  double BAR = 0;
  double BAZ = 0;
  double QUZ = 1;
  for (int i = (1); i <= (n); i++) {
    cin >> PUR;
    FOO = FOO + PUR;
    BAR = BAR + 1;
    if (max(FOO * QUZ, BAR * BAZ) == FOO * QUZ) {
      BAZ = FOO;
      QUZ = BAR;
    }
  }
  printf("%.7lf\n", BAZ / QUZ);
  return 0;
}