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;
long long multiply(long long num1, long long num2, long long mod) {
  long long ans = 0;
  num1 %= mod;
  num2 %= mod;
  while (num2) {
    if (num2 & 1) {
      ans += num1;
      ans %= mod;
    }
    num1 *= 2;
    num1 %= mod;
    num2 /= 2;
  }
  return ans;
}
long long powmod2(long long base, long long exponent, long long mod) {
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) {
      ans = multiply(ans, base, mod);
    }
    base = multiply(base, base, mod);
    exponent /= 2;
  }
  return ans;
}
long long powmod(long long base, long long exponent, long long mod) {
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    exponent /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
const int upperlimit = 1e6 + 100;
const int mod = 1e9 + 7;
int arr[upperlimit];
int main() {
  int n, i, j, k, q, x, v;
  long long sum = 0, tot;
  scanf("%d", &n);
  scanf("%d", &q);
  tot = (1 << n);
  for (i = 0; i < tot; i++) {
    scanf("%d", &arr[i]);
    sum += arr[i];
  }
  double answer = (1.0 * sum) / (1.0 * tot);
  printf("%.10f\n", answer);
  while (q--) {
    scanf("%d", &x);
    scanf("%d", &v);
    sum -= arr[x];
    sum += v;
    arr[x] = v;
    answer = (1.0 * sum) / (1.0 * tot);
    printf("%.10f\n", answer);
  }
  return 0;
}