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;
double const EPS = 1.0E-9;
int const MOD = (int)1e9 + 7;
inline int read(char* s) { return scanf("%s", s); }
inline int read(int& x) { return scanf("%d", &(x)); }
inline int read(long long& x) { return scanf("%lld", &x); }
inline int read(double& x) { return scanf("%lf", &x); }
inline int writeln(int x) { return printf("%d\n", x); }
inline int writeln(long long x) { return printf("%lld\n", x); }
inline int writeln(double x) { return printf("%.12f\n", x); }
template <typename... T>
int read(T&... args) {
  using swallow = int[];
  swallow w = {0, (read(args))...};
  return sizeof(w) / sizeof(w[0]) - 1;
}
template <typename T>
int read(std::vector<T>& v) {
  for (T& t : v) read(t);
  return (int)v.size();
}
inline long long gcd(long long x, long long y) {
  return y == 0 ? x : gcd(y, x % y);
}
inline long long lcm(long long x, long long y) {
  return x == 0 && y == 0 ? 0 : x / gcd(x, y) * y;
}
inline long long powmod(long long x, long long n, long long m = MOD) {
  long long r = 1;
  while (n) {
    if (n & 1) r = (r * x) % m;
    x = (x * x) % m;
    n /= 2;
  }
  return r;
}
static int leap(int y) { return y % 4 == 0 && y % 100 != 0 || y % 400 == 0; }
static int const month[2][12] = {
    {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
    {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};
static const int maxn = 1 << 19;
static const int maxm = 1 << 20;
int n, q;
int a[maxn];
long long sum;
static int solve() {
  read(n, q);
  for (int i = 0; i < (1 << n); ++i) {
    read(a[i]);
    sum += a[i];
  }
  writeln((double)sum / (1 << n));
  for (int i = (0), vzq_b = (q); i < vzq_b; ++i) {
    int z, g;
    read(z, g);
    sum += g - a[z];
    a[z] = g;
    writeln((double)sum / (1 << n));
  }
  return 0;
}
int main(int argc, char* argv[]) {
  solve();
  return 0;
}