Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
class IO {
  static const int bufSize = 1 << 18;
  char inBuf[bufSize], outBuf[bufSize];
  char *ip1 = inBuf, *ip2 = inBuf;
  int goodReadBit = 0, op1 = -1, op2 = bufSize - 1;
  template <typename T>
  __inline__ __attribute__((always_inline)) void __RI(T& x) {
    int ch = gc(), neg = 1;
    x = 0;
    for (; !(isdigit(ch) || ch == '-' || ch == EOF); ch = gc())
      ;
    if (ch == EOF) {
      goodReadBit = 0;
      return;
    }
    if (ch == '-') neg = -1, ch = gc();
    for (; isdigit(ch); ch = gc()) x = x * 10 + (ch - 48) * neg;
  }
  template <typename T>
  __inline__ __attribute__((always_inline)) void __RC(T& x) {
    unsigned char ch;
    while (isspace(ch = gc()))
      ;
    x = ch;
  }
  __inline__ __attribute__((always_inline)) void __RS(std::string& x) {
    char ch;
    x.clear();
    for (ch = gc(); isspace(ch); ch = gc())
      ;
    if (ch == EOF) {
      goodReadBit = 0;
      return;
    }
    for (; !isspace(ch); ch = gc()) x.push_back(ch);
  }

 public:
  __inline__ __attribute__((always_inline)) int gc() {
    return ip1 == ip2 && (ip2 = (ip1 = inBuf) + fread(inBuf, 1, bufSize, stdin),
                          ip1 == ip2)
               ? EOF
               : *ip1++;
  }
  __inline__ __attribute__((always_inline)) void R(char& x) { __RC(x); }
  __inline__ __attribute__((always_inline)) void R(unsigned char& x) {
    __RC(x);
  }
  __inline__ __attribute__((always_inline)) void R(std::string& x) { __RS(x); }
  template <typename T1, typename T2>
  __inline__ __attribute__((always_inline)) void R(std::pair<T1, T2>& x) {
    R(x.first), R(x.second);
  }
  template <typename T>
  __inline__ __attribute__((always_inline)) void R(std::vector<T>& x) {
    for (auto it = x.begin(); it != x.end(); ++it) R(*it);
  }
  template <typename T, typename... Args>
  __inline__ __attribute__((always_inline)) void RA(T* a, int n) {
    for (int i = 0; i < n; ++i) R(a[i]);
  }
  template <typename T, typename... Args>
  __inline__ __attribute__((always_inline)) void R(T& x, Args&... args) {
    R(x), R(args...);
  }
  template <typename T, typename... Args>
  __inline__ __attribute__((always_inline)) void RA(T* a, int n, Args... args) {
    for (int i = 0; i < n; ++i) RA(a[i], args...);
  }
  template <typename T>
  __inline__ __attribute__((always_inline)) void R(T& x) {
    static_assert(std::is_integral<T>::value, "Unsupported types");
    if (std::is_integral<T>::value) __RI(x);
  }

 private:
  char space;
  template <typename T>
  __inline__ __attribute__((always_inline)) void __WI(T x) {
    static char buf[64];
    static int len = -1;
    if (x >= 0) {
      do {
        buf[++len] = x % 10 + 48, x /= 10;
      } while (x);
    } else {
      pc('-');
      do {
        buf[++len] = -(x % 10) + 48, x /= 10;
      } while (x);
    }
    while (len >= 0) {
      pc(buf[len]), --len;
    }
  }

 public:
  __inline__ __attribute__((always_inline)) void pc(const char& x) {
    if (op1 == op2) flush();
    outBuf[++op1] = x;
  }
  __inline__ __attribute__((always_inline)) void flush() {
    fwrite(outBuf, 1, op1 + 1, stdout), op1 = -1;
  }
  __inline__ __attribute__((always_inline)) void W(const char& x) { pc(x); }
  __inline__ __attribute__((always_inline)) void W(const char* x) {
    while (*x != '\0') pc(*(x++));
  }
  __inline__ __attribute__((always_inline)) void W(const std::string& x) {
    W(x.c_str());
  }
  template <typename T>
  __inline__ __attribute__((always_inline)) void W(const std::vector<T>& x) {
    for (auto it = x.begin(); it != x.end(); it++) WS(*it);
    WL();
  }
  __inline__ __attribute__((always_inline)) void WL() { W('\n'); }
  template <typename T>
  __inline__ __attribute__((always_inline)) void WL(const T& x) {
    W(x), W('\n');
  }
  __inline__ __attribute__((always_inline)) void WS() { W(' '); }
  template <typename T>
  __inline__ __attribute__((always_inline)) void WS(const T& x) {
    W(x), W(space);
  }
  template <typename T>
  __inline__ __attribute__((always_inline)) void WA(T* a, int n) {
    for (int i = 0; i < n; i++) WS(a[i]);
    WL();
  }
  template <typename T, typename... Args>
  __inline__ __attribute__((always_inline)) void W(const T& x,
                                                   const Args&... args) {
    W(x), W(space), W(args...);
  }
  template <typename... Args>
  __inline__ __attribute__((always_inline)) void WL(const Args&... args) {
    W(args...), W('\n');
  }
  template <typename T, typename... Args>
  __inline__ __attribute__((always_inline)) void WA(T* a, int n, Args... args) {
    for (int i = 0; i < n; i++) WA(a[i], args...);
  }
  template <typename T>
  __inline__ __attribute__((always_inline)) void W(const T& x) {
    static_assert(std::is_integral<T>::value, "Unsupported types");
    if (std::is_integral<T>::value) __WI(x);
  }
  template <typename T>
  __inline__ __attribute__((always_inline)) IO& operator>>(T& x) {
    R(x);
    return (*this);
  }
  template <typename T>
  __inline__ __attribute__((always_inline)) IO& operator<<(const T& x) {
    W(x);
    return (*this);
  }
  IO() { space = ' '; }
} io;
namespace Alphagocc {
decltype(std::chrono::high_resolution_clock::now()) tStart;
};
void TimeBegin() {
  Alphagocc::tStart = std::chrono::high_resolution_clock::now();
}
long long TimeEnd() {
  return (std::chrono::high_resolution_clock::now() - Alphagocc::tStart)
      .count();
}
using namespace std;
const int8_t INFC = 0x3f;
const int16_t INFS = 0x3f3f;
const int32_t INF = 0x3f3f3f3f;
const int64_t INFL = 0x3f3f3f3f3f3f3f3fLL;
int main() {
  int n;
  io >> n;
  vector<int> a(n);
  io >> a;
  vector<pair<double, int>> st;
  for (int i = 0; i < n; i++) {
    st.emplace_back(a[i], 1);
    while (st.size() > 1 &&
           (*(st.end() - 2)).first >= (*(st.end() - 1)).first) {
      (*(st.end() - 2)).first =
          ((*(st.end() - 2)).first * (*(st.end() - 2)).second +
           (*(st.end() - 1)).first * (*(st.end() - 1)).second) /
          ((*(st.end() - 2)).second + (*(st.end() - 1)).second);
      (*(st.end() - 2)).second += (*(st.end() - 1)).second;
      st.pop_back();
    }
  }
  cout << fixed << setprecision(10);
  for (int i = 0; i < st.size(); i++)
    while (st[i].second--) cout << st[i].first << "\n";
  cout << endl;
  return 0;
}