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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        EyeClosed solver = new EyeClosed();
        solver.solve(1, in, out);
        out.close();
    }

    static class EyeClosed {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int N = in.nextInt();
            int q = in.nextInt();
            SegmentTree seg = new SegmentTree(N);
            for (int i = 0; i < N; i++) {
                seg.add(1, 0, N - 1, i, i, in.nextInt());
            }
            while (q-- > 0) {
                int id = in.nextInt();
                if (id == 1) {
                    int l1 = in.nextInt() - 1;
                    int r1 = in.nextInt() - 1;
                    int l2 = in.nextInt() - 1;
                    int r2 = in.nextInt() - 1;
                    long length1 = (r1 - l1 + 1);
                    long length2 = (r2 - l2 + 1);
                    double sum1 = seg.query(1, 0, N - 1, l1, r1);
                    double sum2 = seg.query(1, 0, N - 1, l2, r2);
                    seg.mult(1, 0, N - 1, l1, r1, 1 - (1.0) / (length1));
                    seg.mult(1, 0, N - 1, l2, r2, 1 - (1.0) / (length2));
                    seg.add(1, 0, N - 1, l1, r1, sum2 / ((length1) * (length2)));
                    seg.add(1, 0, N - 1, l2, r2, sum1 / ((length2) * (length1)));
                } else {
                    out.println(seg.query(1, 0, N - 1, in.nextInt() - 1, in.nextInt() - 1));
                }
            }
        }

        class SegmentTree {
            double[] sum;
            double[] lzyMult;
            double[] lzyAdd;

            SegmentTree(int N) {
                sum = new double[4 * N];
                lzyMult = new double[4 * N];
                lzyAdd = new double[4 * N];
                Arrays.fill(lzyMult, 1);
            }

            void add(int n, int tl, int tr, int l, int r, double val) {
                if (tl > r || tr < l) {
                    return;
                }
                if (tl >= l && tr <= r) {
                    lzyAdd[n] += val;
                    sum[n] += (tr - tl + 1) * val;
                } else {
                    push(n, tl, tr);
                    int tm = (tl + tr) / 2;
                    add(2 * n, tl, tm, l, r, val);
                    add(2 * n + 1, tm + 1, tr, l, r, val);
                    sum[n] = sum[2 * n + 1] + sum[2 * n];
                }
            }

            void mult(int n, int tl, int tr, int l, int r, double val) {
                if (tl > r || tr < l) {
                    return;
                }
                if (tl >= l && tr <= r) {
                    lzyMult[n] *= val;
                    lzyAdd[n] *= val;
                    sum[n] *= val;
                } else {
                    push(n, tl, tr);
                    int tm = (tl + tr) / 2;
                    mult(2 * n, tl, tm, l, r, val);
                    mult(2 * n + 1, tm + 1, tr, l, r, val);
                    sum[n] = sum[2 * n] + sum[2 * n + 1];
                }
            }

            double query(int n, int tl, int tr, int l, int r) {
                if (tl > r || tr < l) {
                    return 0;
                }
                if (tl >= l && tr <= r) {
                    return sum[n];
                } else {
                    push(n, tl, tr);
                    int tm = (tl + tr) / 2;
                    return query(2 * n, tl, tm, l, r) + query(2 * n + 1, tm + 1, tr, l, r);
                }
            }

            void push(int n, int tl, int tr) {
                int tm = (tl + tr) / 2;
                sum[2 * n] *= lzyMult[n];
                sum[2 * n + 1] *= lzyMult[n];
                sum[2 * n] += lzyAdd[n] * (tm - tl + 1);
                sum[2 * n + 1] += lzyAdd[n] * (tr - (tm + 1) + 1);
                lzyAdd[2 * n] *= lzyMult[n];
                lzyAdd[2 * n + 1] *= lzyMult[n];
                lzyMult[2 * n] *= lzyMult[n];
                lzyMult[2 * n + 1] *= lzyMult[n];
                lzyAdd[2 * n] += lzyAdd[n];
                lzyAdd[2 * n + 1] += lzyAdd[n];
                lzyMult[n] = 1;
                lzyAdd[n] = 0;
            }

        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}