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;
class Solution {
 public:
  Solution(vector<int>& A) : nums(A), sum(0), expectedVal(0) {
    for (auto num : nums) {
      sum += num;
    }
    expectedVal = (0.0 + sum) / nums.size();
  }
  double query(int z, int g) {
    sum += g - nums[z];
    nums[z] = g;
    expectedVal = (0.0 + sum) / nums.size();
    return expectedVal;
  }
  vector<double> query(vector<pair<int, int>>& qs) {
    vector<double> res;
    res.push_back(expectedVal);
    for (auto& q : qs) {
      res.push_back(query(q.first, q.second));
    }
    return res;
  }
  double get() { return expectedVal; }

 private:
  vector<int> nums;
  long long sum;
  double expectedVal;
};
int main(int argc, char** argv) {
  int n, r;
  scanf("%d%d", &n, &r);
  int N = 1 << n;
  vector<int> nums(N);
  for (int i = 0; i < N; ++i) {
    scanf("%d", &nums[i]);
  }
  Solution sol(nums);
  vector<pair<int, int>> qs;
  qs.reserve(r);
  for (int i = 0, z = 0, g = 0; i < r; ++i) {
    scanf("%d%d", &z, &g);
    qs.emplace_back(z, g);
  }
  auto res = sol.query(qs);
  for (auto r : res) {
    printf("%.12lf\n", r);
  }
  return 0;
}