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;
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-ffloat-store")
#pragma GCC optimize("-fno-defer-pop")
long long dx[4] = {-1, 0, 0, +1};
long long dy[4] = {0, -1, +1, 0};
const long long LINF = 1e18;
const long long INF = 1e9;
const long long MOD = 1e9 + 7;
long long strtoll(string s) {
  stringstream str(s);
  long long number = 0;
  str >> number;
  return number;
}
string lltostr(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
long long llsqrt(long long x) {
  if (x == 0 || x == 1) return x;
  long long start = 1, end = x / 2, ans;
  while (start <= end) {
    long long mid = (start + end) / 2;
    if (mid * mid == x)
      return mid;
    else if (mid <= x / mid)
      ans = mid, start = mid + 1;
    else
      end = mid - 1;
  }
  return ans;
}
vector<long long> spf;
void sieve(long long size) {
  spf.resize(size + 1);
  for (long long i = 1; i <= size; ++i) spf[i] = i;
  for (long long i = 4; i <= size; i += 2) spf[i] = 2;
  for (long long i = 3; i * i <= size; ++i) {
    if (spf[i] == i) {
      for (long long j = i * i; j <= size; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
long long power(long long a, long long b, long long m) {
  if (b == 0) return 1;
  if (b == 1) return a % m;
  long long t = power(a, b / 2, m);
  t = (t * t) % m;
  if (b & 1) t = (t * a) % m;
  return t;
}
long long modInverse(long long a, long long m) { return power(a, m - 2, m); }
void fastscan(long long &number) {
  bool negative = false;
  register long long c;
  number = 0;
  c = getchar();
  if (c == '-') {
    negative = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) number = number * 10 + c - 48;
  if (negative) number *= -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, r;
  cin >> n >> r;
  long long f[1 << n];
  for (long long i = 0; i < (1 << n); i++) cin >> f[i];
  long long s = 0;
  for (long long i = 0; i < (1 << n); i++) s += f[i];
  double tot = (1 << n);
  printf("%.8lf\n", 1.0 * s / (1 << n));
  for (long long i = 0; i < r; i++) {
    long long x, g;
    cin >> x >> g;
    long long d = g - f[x];
    s += d;
    f[x] = g;
    printf("%.8lf\n", s / tot);
  }
  return 0;
}