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;
vector<pair<int, double>> V[2 << 18];
int main() {
  int n, r;
  scanf("%d%d", &n, &r);
  int i, j;
  for (i = (0); i < (1 << n); ++i) {
    int a;
    scanf("%d", &a);
    V[i].push_back(pair<int, int>(0, a));
  }
  for (i = (0); i < (r); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    V[a].push_back(make_pair(i + 1, b));
  }
  int I[32];
  I[0] = 0;
  for (i = (0); i < (n); ++i) I[i + 1] = I[i] + (1 << (n - i));
  for (i = (1); i < (n + 1); ++i) {
    int len = 1 << (n - i);
    for (j = (0); j < (len); ++j) {
      int p = I[i] + j;
      int a = I[i - 1] + j + j;
      int b = I[i - 1] + j + j + 1;
      int ia = 0;
      int ib = 0;
      while (ia < int((V[a]).size()) || ib < int((V[b]).size())) {
        int pos = (ib == int((V[b]).size()) ||
                   (ia < int((V[a]).size()) && V[a][ia].first < V[b][ib].first))
                      ? V[a][ia].first
                      : V[b][ib].first;
        if (ia != int((V[a]).size()) && V[a][ia].first <= pos) ++ia;
        if (ib != int((V[b]).size()) && V[b][ib].first <= pos) ++ib;
        V[p].push_back(
            make_pair(pos, (V[a][ia - 1].second + V[b][ib - 1].second) / 2));
      }
    }
  }
  int a = I[n];
  int pos = 0;
  for (i = (0); i < (r + 1); ++i) {
    if (pos != int((V[a]).size()) && V[a][pos].first <= i) ++pos;
    printf("%.11lf\n", V[a][pos - 1].second);
  }
  return 0;
};