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 JAVA):

Solution:
//package round492;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class D {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		int n = ni(), r = ni();
		long[] s = new long[1<<n];
		long sum = 0;
		for(int i = 0;i < 1<<n;i++){
			s[i] = nl();
			sum += s[i];
		}
		out.println(dtos((double)sum/(1<<n), 14));
		for(int i = 0;i < r;i++){
			int w = ni(), g = ni();
			sum += g - s[w];
			s[w] = g;
			out.println(dtos((double)sum/(1<<n), 14));
		}
	}
	
	
	
	public static String dtos(double x, int n) {
		StringBuilder sb = new StringBuilder();
		if(x < 0){
			sb.append('-');
			x = -x;
		}
		x += Math.pow(10, -n)/2;
//		if(x < 0){ x = 0; }
		sb.append((long)x);
		sb.append(".");
		x -= (long)x;
		for(int i = 0;i < n;i++){
			x *= 10;
			sb.append((int)x);
			x -= (int)x;
		}
		return sb.toString();
	}

	
	void run() throws Exception
	{
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception { new D().run(); }
	
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;
	
	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	
	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}