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;
template <class T>
using min_queue = priority_queue<T, vector<T>, greater<T>>;
template <typename Args>
void kill(Args args) {
  cout << args << "\n";
  exit(0);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const double PI = acos(-1);
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const int N = 1 << 19;
long double arr[N];
long double mul = 1;
int n, r;
void solve(int testnum) {
  cin >> n >> r;
  for (int i = 1; i <= n; i++) {
    mul *= 0.5;
  }
  long double sum = 0;
  for (int i = 0; i < (1 << n); i++) {
    cin >> arr[i];
    sum += arr[i];
  }
  cout << sum * mul << "\n";
  for (int i = 1; i <= r; i++) {
    int g;
    long double x;
    cin >> g >> x;
    sum -= arr[g];
    sum += x;
    arr[g] = x;
    cout << mul * sum << "\n";
  }
}
void precompute() {}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  int testcase = 1;
  clock_t zzzx = clock();
  precompute();
  for (int i = 1; i <= testcase; i++) {
    solve(i);
  }
  double elapsed_time = (double)(clock() - zzzx) / CLOCKS_PER_SEC;
  cerr << "elapsed_time"
       << " = " << (elapsed_time) << "\n";
  return 0;
}