      
    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;
const long long N = 1e5;
long long a[N], used[N], con[N];
vector<long long> g[N];
inline void read(long long &a) {
  char c;
  bool neg = 0;
  c = getchar();
  a = 0;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    a = (a << 3) + (a << 1) + (c - '0');
    c = getchar();
  }
  if (neg) a *= -1;
}
inline bool dfs(long long u, long long now) {
  if (used[u] == now) {
    return 0;
  }
  used[u] = now;
  for (auto to : g[u]) {
    if (con[to] == -1 || dfs(con[to], now) == 1) {
      con[to] = u;
      return 1;
    }
  }
  return 0;
}
inline void solve() {
  long long n;
  cin >> n;
  long double now = 0, maxx = 0;
  for (long long i = 0; i < n; i++) {
    long long a;
    cin >> a;
    now += a;
    maxx = max(maxx, now / (i + 1));
  }
  cout << setprecision(20) << maxx;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
