Problem: Allen and Bessie are playing a simple number game. They both know a function f: \{0, 1\}^n → R, i. e. the function takes n binary arguments and returns a real value. At the start of the game, the variables x_1, x_2, ..., x_n are all set to -1. Each round, with equal probability, one of Allen or Bessie gets to make a move. A move consists of picking an i such that x_i = -1 and either setting x_i → 0 or x_i → 1.

After n rounds all variables are set, and the game value resolves to f(x_1, x_2, ..., x_n). Allen wants to maximize the game value, and Bessie wants to minimize it.

Your goal is to help Allen and Bessie find the expected game value! They will play r+1 times though, so between each game, exactly one value of f changes. In other words, between rounds i and i+1 for 1 ≤ i ≤ r, f(z_1, ..., z_n) → g_i for some (z_1, ..., z_n) ∈ \{0, 1\}^n. You are to find the expected game value in the beginning and after each change.

Input

The first line contains two integers n and r (1 ≤ n ≤ 18, 0 ≤ r ≤ 2^{18}).

The next line contains 2^n integers c_0, c_1, ..., c_{2^n-1} (0 ≤ c_i ≤ 10^9), denoting the initial values of f. More specifically, f(x_0, x_1, ..., x_{n-1}) = c_x, if x = \overline{x_{n-1} … x_0} in binary.

Each of the next r lines contains two integers z and g (0 ≤ z ≤ 2^n - 1, 0 ≤ g ≤ 10^9). If z = \overline{z_{n-1} ... z_0} in binary, then this means to set f(z_0, ..., z_{n-1}) → g.

Output

Print r+1 lines, the i-th of which denotes the value of the game f during the i-th round. Your answer must have absolute or relative error within 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.

Examples

Input

2 2
0 1 2 3
2 5
0 4


Output

1.500000
2.250000
3.250000


Input

1 0
2 3


Output

2.500000


Input

2 0
1 1 1 1


Output

1.000000

Note

Consider the second test case. If Allen goes first, he will set x_1 → 1, so the final value will be 3. If Bessie goes first, then she will set x_1 → 0 so the final value will be 2. Thus the answer is 2.5.

In the third test case, the game value will always be 1 regardless of Allen and Bessie's play.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
clock_t timeStart, timeFinish;
void timeBegin() { timeStart = clock(); }
void timeEnd() { timeFinish = clock(); }
void timeDuration() {
  timeEnd();
  double time_taken = double(timeFinish - timeStart) / double(CLOCKS_PER_SEC);
  cout << "Time taken by program is : " << fixed << time_taken
       << setprecision(5);
  cout << " sec " << endl;
}
class InputReader {
 public:
  InputReader() {
    input_file = stdin;
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  InputReader(const char *file_name) {
    input_file = fopen(file_name, "r");
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  inline InputReader &operator>>(int &n) {
    while ((buffer[cursor] < '0' || buffer[cursor] > '9') &&
           buffer[cursor] != '-') {
      advance();
    }
    int sign = 1;
    if (buffer[cursor] == '-') {
      sign = -1;
      advance();
    }
    n = 0;
    while ('0' <= buffer[cursor] && buffer[cursor] <= '9') {
      n = n * 10 + buffer[cursor] - '0';
      advance();
    }
    n *= sign;
    return *this;
  }

 private:
  FILE *input_file;
  static const int SIZE = 1 << 17;
  int cursor;
  char buffer[SIZE];
  inline void advance() {
    ++cursor;
    if (cursor == SIZE) {
      cursor = 0;
      fread(buffer, SIZE, 1, input_file);
    }
  }
};
const int MAXN = 18;
int v[1 << MAXN];
int main() {
  timeBegin();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  long long sum = 0;
  for (int x = 0; x < (1 << n); x++) {
    cin >> v[x];
    sum += v[x];
  }
  cout << fixed << setprecision(8) << (long double)sum / (1 << n) << "\n";
  for (int i = 1; i <= q; i++) {
    int x, val;
    cin >> x >> val;
    sum -= v[x];
    v[x] = val;
    sum += v[x];
    cout << fixed << setprecision(8) << (long double)sum / (1 << n) << "\n";
  }
  return 0;
}