There are n water tanks in a row, i-th of them contains a_i liters of water. The tanks are numbered from 1 to n from left to right.

You can perform the following operation: choose some subsegment [l, r] (1≤ l ≤ r ≤ n), and redistribute water in tanks l, l+1, ..., r evenly. In other words, replace each of a_l, a_{l+1}, ..., a_r by \frac{a_l + a_{l+1} + ... + a_r}{r-l+1}. For example, if for volumes [1, 3, 6, 7] you choose l = 2, r = 3, new volumes of water will be [1, 4.5, 4.5, 7]. You can perform this operation any number of times.

What is the lexicographically smallest sequence of volumes of water that you can achieve?

As a reminder:

A sequence a is lexicographically smaller than a sequence b of the same length if and only if the following holds: in the first (leftmost) position where a and b differ, the sequence a has a smaller element than the corresponding element in b.

Input

The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of water tanks.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — initial volumes of water in the water tanks, in liters.

Because of large input, reading input as doubles is not recommended.

Output

Print the lexicographically smallest sequence you can get. In the i-th line print the final volume of water in the i-th tank.

Your answer is considered correct if the absolute or relative error of each a_i does not exceed 10^{-9}.

Formally, let your answer be a_1, a_2, ..., a_n, and the jury's answer be b_1, b_2, ..., b_n. Your answer is accepted if and only if \frac{|a_i - b_i|}{max{(1, |b_i|)}} ≤ 10^{-9} for each i.

Examples

Input


4
7 5 5 7


Output


5.666666667
5.666666667
5.666666667
7.000000000


Input


5
7 8 8 10 12


Output


7.000000000
8.000000000
8.000000000
10.000000000
12.000000000


Input


10
3 9 5 5 1 7 5 3 8 7


Output


3.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
7.500000000
7.500000000

Note

In the first sample, you can get the sequence by applying the operation for subsegment [1, 3].

In the second sample, you can't get any lexicographically smaller sequence.
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using ld = long double;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vpi = vector<pi>;
using si = set<int>;
using sl = set<ll>;
using qi = queue<int>;
using ql = queue<ll>;
template <class T>
bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
template <class T>
bool uin(T& u, T& v, T a, T b) {
  return v - u > b - a ? (u = a, v = b, true) : false;
}
template <class T>
bool uax(T& u, T& v, T a, T b) {
  return v - u < b - a ? (u = a, v = b, true) : false;
}
namespace input {
template <class T>
void re(complex<T>& x);
template <class T1, class T2>
void re(pair<T1, T2>& p);
template <class T>
void re(vector<T>& a);
template <class T, size_t SZ>
void re(array<T, SZ>& a);
template <class T>
void re(T& x) {
  cin >> x;
}
void re(double& x) {
  string t;
  re(t);
  x = stod(t);
}
void re(ld& x) {
  string t;
  re(t);
  x = stold(t);
}
template <class T, class... Ts>
void re(T& t, Ts&... ts) {
  re(t);
  re(ts...);
}
template <class T>
void re(complex<T>& x) {
  T a, b;
  re(a, b);
  x = cd(a, b);
}
template <class T1, class T2>
void re(pair<T1, T2>& p) {
  re(p.f, p.s);
}
template <class T>
void re(vector<T>& a) {
  for (int i = 0; i < (((int)(a).size())); ++i) re(a[i]);
}
template <class T, size_t SZ>
void re(array<T, SZ>& a) {
  for (int i = 0; i < (SZ); ++i) re(a[i]);
}
}  // namespace input
using namespace input;
namespace output {
void pr(int x) { cout << x; }
void pr(long x) { cout << x; }
void pr(ll x) { cout << x; }
void pr(unsigned x) { cout << x; }
void pr(unsigned long x) { cout << x; }
void pr(unsigned long long x) { cout << x; }
void pr(float x) { cout << x; }
void pr(double x) { cout << x; }
void pr(ld x) { cout << x; }
void pr(char x) { cout << x; }
void pr(const char* x) { cout << x; }
void pr(const string& x) { cout << x; }
void pr(bool x) { pr(x ? "true" : "false"); }
template <class T>
void pr(const complex<T>& x) {
  cout << x;
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T>
void pr(const T& x);
template <class T, class... Ts>
void pr(const T& t, const Ts&... ts) {
  pr(t);
  pr(ts...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr("{", x.f, ", ", x.s, "}");
}
template <class T>
void pr(const T& x) {
  pr("{");
  bool fst = 1;
  for (const auto& a : x) pr(!fst ? ", " : "", a), fst = 0;
  pr("}");
}
void ps() { pr("\n"); }
template <class T, class... Ts>
void ps(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) pr(" ");
  ps(ts...);
}
void pc() { pr("]\n"); }
template <class T, class... Ts>
void pc(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) pr(", ");
  pc(ts...);
}
}  // namespace output
using namespace output;
inline int bs(bool (*valid)(int), int l, int r, bool order) {
  l--, r++;
  if (!order) swap(l, r);
  while (abs(r - l) > 1) {
    int mid = (l + r) >> 1;
    if (valid(mid))
      r = mid;
    else
      l = mid;
  }
  valid(r);
  return r;
}
struct dsu {
  vector<int> p;
  dsu(int n) { p.resize(n + 1); }
  inline int get(int x) { return p[x] ? p[x] = get(p[x]) : x; }
  inline bool mrg(int x, int y) {
    return get(x) == get(y) ? false : p[get(x)] = get(y);
  }
};
const int M = 1e9 + 7;
const ll lnf = 1e18 + 3;
const int inf = 1e9 + 3;
const int nax = 1e6 + 5;
typedef decay<decltype(M)>::type T;
struct mi {
  T val;
  explicit operator T() const { return val; }
  mi() { val = 0; }
  mi(const ll& v) {
    val = (-M <= v && v <= M) ? v : v % M;
    if (val < 0) val += M;
  }
  friend void pr(const mi& a) { pr(a.val); }
  friend void re(mi& a) {
    ll x;
    re(x);
    a = mi(x);
  }
  friend bool operator==(const mi& a, const mi& b) { return a.val == b.val; }
  friend bool operator!=(const mi& a, const mi& b) { return !(a == b); }
  friend bool operator<(const mi& a, const mi& b) { return a.val < b.val; }
  mi operator-() const { return mi(-val); }
  mi& operator+=(const mi& m) {
    if ((val += m.val) >= M) val -= M;
    return *this;
  }
  mi& operator-=(const mi& m) {
    if ((val -= m.val) < 0) val += M;
    return *this;
  }
  mi& operator*=(const mi& m) {
    val = (ll)val * m.val % M;
    return *this;
  }
  friend mi pow(mi a, ll p) {
    mi ans = 1;
    assert(p >= 0);
    for (; p; p /= 2, a *= a)
      if (p & 1) ans *= a;
    return ans;
  }
  friend mi inv(const mi& a) {
    assert(a != 0);
    return pow(a, M - 2);
  }
  mi& operator/=(const mi& m) { return (*this) *= inv(m); }
  friend mi operator+(mi a, const mi& b) { return a += b; }
  friend mi operator-(mi a, const mi& b) { return a -= b; }
  friend mi operator*(mi a, const mi& b) { return a *= b; }
  friend mi operator/(mi a, const mi& b) { return a /= b; }
};
vi invs, fac, ifac;
void genFac(int SZ) {
  invs.resize(SZ), fac.resize(SZ), ifac.resize(SZ);
  invs[1] = fac[0] = ifac[0] = 1;
  for (int i = (2); i <= (SZ - 1); ++i)
    invs[i] = M - (ll)M / i * invs[M % i] % M;
  for (int i = (1); i <= (SZ - 1); ++i) {
    fac[i] = (ll)fac[i - 1] * i % M;
    ifac[i] = (ll)ifac[i - 1] * invs[i] % M;
  }
}
ll comb(int a, int b) {
  if (a < b || b < 0) return 0;
  return (ll)fac[a] * ifac[b] % M * ifac[a - b] % M;
}
ll p[nax], s, c;
int t, n, q[nax];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  re(n);
  vl a(n);
  re(a);
  for (int i = 0; i < (n); ++i) {
    s = a[i], c = 1;
    while (t && p[t] * c >= s * q[t]) s += p[t], c += q[t--];
    p[++t] = s, q[t] = c;
  }
  cout << setprecision(10);
  for (int i = (1); i <= (t); ++i)
    for (int j = 0; j < (q[i]); ++j) cout << (double)p[i] / q[i] << '\n';
}
