Problem: Vasya and Petya were tired of studying so they decided to play a game. Before the game begins Vasya looks at array a consisting of n integers. As soon as he remembers all elements of a the game begins. Vasya closes his eyes and Petya does q actions of one of two types:

1) Petya says 4 integers l1, r1, l2, r2 — boundaries of two non-intersecting segments. After that he swaps one random element from the [l1, r1] segment with another random element from the [l2, r2] segment.

2) Petya asks Vasya the sum of the elements of a in the [l, r] segment.

Vasya is a mathematician so he answers Petya the mathematical expectation of the sum of the elements in the segment.

Your task is to write a program which will answer the second type questions as Vasya would do it. In other words your program should print the mathematical expectation of the sum of the elements of a in the [l, r] segment for every second type query.

Input

The first line contains two integers n, q (2 ≤ n ≤ 105, 1 ≤ q ≤ 105) — the number of elements in the array and the number of queries you need to handle.

The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of the array.

The next q lines contain Petya's actions of type 1 or 2.

If it is a type 1 action then the line contains 5 integers 1, l1, r1, l2, r2 (1 ≤ l1 ≤ r1 ≤ n, 1 ≤ l2 ≤ r2 ≤ n).

If it is a type 2 query then the line contains 3 integers 2, l, r (1 ≤ l ≤ r ≤ n).

It is guaranteed that there is at least one type 2 query and segments [l1, r1], [l2, r2] don't have common elements. 

Output

For each type 2 query print one real number — the mathematical expectation of the sum of elements in the segment.

Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 4 — formally, the answer is correct if <image> where x is jury's answer and y is yours.

Examples

Input

4 4
1 1 2 2
1 2 2 3 3
2 1 2
1 1 2 3 4
2 1 2


Output

3.0000000
3.0000000


Input

10 5
1 1 1 1 1 2 2 2 2 2
1 1 5 6 10
2 1 5
1 1 5 6 10
1 1 5 6 10
2 6 10


Output

6.0000000
8.0400000


Input

10 10
1 2 3 4 5 6 7 8 9 10
1 1 5 6 10
1 1 5 6 10
2 1 5
1 1 3 6 9
2 1 3
1 5 7 8 10
1 1 1 10 10
2 1 5
2 7 10
2 1 10


Output

23.0000000
14.0000000
28.0133333
21.5733333
55.0000000

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class _895E_3 {

	public static void main(String args[]) {
		FastScanner in = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int n = in.nextInt();
		int q = in.nextInt();
		sTree st = new sTree(n);

		double a[] = new double[n];
		for(int i = 0; i < n; i++)
			a[i] = in.nextDouble();
		 
		//long start = System.currentTimeMillis();
		st.fillTree(a);

		
		for(int i = 0; i < q; i++) {
			int type = in.nextInt();
			if(type == 1) {
				int l1 = in.nextInt()-1;
				int r1 = in.nextInt()-1;
				int l2 = in.nextInt()-1;
				int r2 = in.nextInt()-1;
				st.rangeSwap(l1, r1, l2, r2);
			}
			else {
				int l1 = in.nextInt()-1;
				int r1 = in.nextInt()-1;
				out.println(st.rangeSum(1, l1, r1));
			}
		}
		//long end = System.currentTimeMillis();
		//out.println(end-start+" millis");
		
		out.close();
	}
	
	
	static class sTree{
		int lo[];
		int hi[];
		double prod[];
		double add[];

		double sum[];
		double childs[];
		int tot;
		
		public sTree(int n)//Create tree of size n
		{
			if(Integer.bitCount(n) == 1) tot = 2*n;
			else tot =2*(Integer.highestOneBit(n) << 1);
			lo = new int[tot+1];
			hi = new int[tot+1];
			childs = new double[tot+1];
			prod = new double[tot+1];
			add = new double[tot+1];

			sum = new double[tot+1];
			Arrays.fill(prod, 1);
			Arrays.fill(add, 0);

			
			makeTree(1, 0, (tot/2)-1);
		}
		public void makeTree(int i, int left, int right)
		{
			lo[i] = left;
			hi[i] = right;
			int c1 = i*2;
			int c2 = i*2+1;
			if(c2 >= tot) return;
			int mid = (left+right)/2;
			makeTree(c1, left, mid);
			makeTree(c2, mid+1, right);
		}
		public void fillTree(double a[]) {
			int ind = tot/2;
			for(int i = 0; i < a.length; i++) {
				sum[ind+i] = a[i];
			}
			init(1);
			init2(1);
		}
		public double init(int ind) {
			int c1 = ind*2;
			int c2 = ind*2+1;
			if(c1 >= tot) return sum[ind];
			else return sum[ind] = (init(c1)+init(c2));
		}
		public double init2(int ind) {
			int c1 = ind*2;
			int c2 = ind*2+1;
			if(c1 >= tot) return childs[ind] = 1;
			else return childs[ind] = init2(c1)+init2(c2);
		}
		
		
		public void rangeSwap(int l1, int r1, int l2, int r2) {
			double avg1 = rangeSum(1, l1, r1)/(r1-l1+1);
			double avg2 = rangeSum(1, l2, r2)/(r2-l2+1);
			//if(Math.abs(avg1-avg2) < .000001) return;
			//System.out.println(avg1+" "+avg2);
			
			double m1 = (double)(r1-l1)/(double)(r1-l1+1);
			double add1 = avg2/(r1-l1+1);
			
			double m2 = (double)(r2-l2)/(double)(r2-l2+1);
			double add2 = avg1/(r2-l2+1);
			add(1, l1, r1, m1, add1);
			add(1, l2, r2, m2, add2);
		}
		
		public void add(int i, int low, int high, double m, double a)
		{
			if(low <= lo[i] && high >= hi[i]){
				prod[i] *= m;
				add[i] = add[i]*m + a;
				return;
			}
			if(low > hi[i] || high < lo[i]) return;
			prop(i);
			add(i*2, low, high, m, a);
			add(i*2+1, low, high, m, a);
			update(i);
		}
		
		public double rangeSum(int i, int left, int right) {
			if(hi[i] < left || right < lo[i]) 
				return 0;
			if(left <= lo[i] && hi[i] <= right)
				return getSum(i);
			prop(i);
			double sLeft = rangeSum(2*i, left, right);
			double sRight=  rangeSum(2*i+1, left, right);
			update(i);
			return sLeft+sRight;
		}

		public double getSum(int i) {
			return ((sum[i]/childs[i])*prod[i]+add[i]) * childs[i];
		}
		public void prop(int i) {
			if(i*2+1 >= tot) return;
			prod[i*2] *= prod[i];
			add[i*2] = add[i*2]*prod[i] + add[i];
			
			prod[i*2+1] *= prod[i];
			add[i*2+1] = add[i*2+1]*prod[i] + add[i];
			
			prod[i] = 1;
			add[i] = 0;
		}
		
		public void update(int i) {
			sum[i] = getSum(i*2)+getSum(i*2+1);
		}
		
	}
	
	
	static class FastScanner{
		BufferedReader br;
		StringTokenizer st;
		public FastScanner(String s) {
			try{
				br = new BufferedReader(new FileReader(s));
			}
			catch(FileNotFoundException e) {
				e.printStackTrace();
			}
		}
		public FastScanner(){
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String nextToken()	{
			while(st == null ||!st.hasMoreElements()){
				try {
				st = new StringTokenizer(br.readLine());}
				catch(IOException e) {
					e.printStackTrace();
				}			
			}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(nextToken());
		}
		long nextLong() {
			return Long.parseLong(nextToken());
		}
		double nextDouble() {
			return Double.parseDouble(nextToken());
		}
		String next() {
			return nextToken();
		}
		
	}
}