      
    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;
template <class T>
void pp(T v) {
  for (__typeof((v).begin()) it = (v).begin(); it != (v).end(); ++it)
    cout << *it << ' ';
  cout << endl;
}
template <class T>
void pp(T v, int n) {
  for (int i = 0; i < (int)n; i++) cout << v[i] << ' ';
  cout << endl;
}
template <class T>
inline void chmax(T &a, const T b) {
  a = max(a, b);
}
template <class T>
inline void chmin(T &a, const T b) {
  a = min(a, b);
}
const int INF = 1 << 28;
const double EPS = 1.0e-9;
static const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
int main(void) {
  int TUX;
  cin >> TUX;
  int FOO = 0;
  int BAR = 0;
  int BAZ = 0;
  int QUZ = 1;
  while (--TUX >= 0) {
    int PUR;
    cin >> PUR;
    FOO = FOO + PUR;
    BAR = BAR + 1;
    if (FOO * QUZ < BAR * BAZ) break;
    BAZ = FOO;
    QUZ = BAR;
  }
  double ans = (double)BAZ / (double)QUZ;
  printf("%0.8f\n", ans);
  return 0;
}
