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.StringTokenizer;
import java.text.DecimalFormat;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.text.NumberFormat;
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);
        E_Eyes_Closed solver = new E_Eyes_Closed();
        solver.solve(1, in, out);
        out.close();
    }

    static class E_Eyes_Closed {
        final int MAXN = (int) (1e6 + 4);
        int n;
        int q;
        double[] a = new double[MAXN];
        double[] add = new double[MAXN];
        double[] mult = new double[MAXN];
        double[] t = new double[MAXN];

        void build(int v, int tl, int tr) {
            if (tl == tr) {
                t[v] = a[tl];
                mult[v] = 1;
                return;
            }
            int tm = tl + (tr - tl >> 1);
            build(v << 1, tl, tm);
            build(v << 1 | 1, tm + 1, tr);
            t[v] = t[v << 1] + t[v << 1 | 1];
            mult[v] = 1;
        }

        void push(int v, int tl, int tr) {
            int ls = v << 1, rs = v << 1 | 1;
            t[ls] *= mult[v];
            t[rs] *= mult[v];
            mult[ls] *= mult[v];
            mult[rs] *= mult[v];
            add[ls] *= mult[v];
            add[rs] *= mult[v];
            mult[v] = 1;
            int tm = tl + (tr - tl >> 1);
            t[ls] += add[v] * (tm - tl + 1);
            t[rs] += add[v] * (tr - tm);
            add[ls] += add[v];
            add[rs] += add[v];
            add[v] = 0;
            t[v] = t[ls] + t[rs];
        }

        void update(int v, int tl, int tr, int l, int r, double mult_val, double add_val) {
            if (l > r) return;
            if (tl == l && tr == r) {
                t[v] *= mult_val;
                t[v] += add_val * (tr - tl + 1);
                mult[v] *= mult_val;
                add[v] *= mult_val;
                add[v] += add_val;
                return;
            }
            push(v, tl, tr);
            int tm = tl + (tr - tl >> 1);
            update(v << 1, tl, tm, l, Math.min(tm, r), mult_val, add_val);
            update(v << 1 | 1, tm + 1, tr, Integer.max(l, tm + 1), r, mult_val, add_val);
            t[v] = t[v << 1] + t[v << 1 | 1];
        }

        double get(int v, int tl, int tr, int l, int r) {
            if (tl == l && tr == r)
                return t[v];
            if (l > r)
                return 0;
            push(v, tl, tr);
            int tm = tl + (tr - tl >> 1);
            return get(v << 1, tl, tm, l, Math.min(r, tm)) +
                    get(v << 1 | 1, tm + 1, tr, Integer.max(l, tm + 1), r);
        }

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            n = in.nextInt();
            q = in.nextInt();
            for (int i = 0; i < n; ++i)
                a[i] = in.nextInt();
            build(1, 0, n - 1);
            while (q != 0) {
                int type;
                type = in.nextInt();
                if (type == 1) {
                    int l1 = in.nextInt();
                    int r1 = in.nextInt();
                    int l2 = in.nextInt();
                    int r2 = in.nextInt();
                    l1--;
                    r1--;
                    l2--;
                    r2--;
                    double avg_val1 = get(1, 0, n - 1, l1, r1) / (double) (r1 - l1 + 1);
                    double avg_val2 = get(1, 0, n - 1, l2, r2) / (double) (r2 - l2 + 1);
                    update(1, 0, n - 1, l1, r1, 1.0 * (r1 - l1) / (r1 - l1 + 1), avg_val2 / (r1 - l1 + 1));
                    update(1, 0, n - 1, l2, r2, 1.0 * (r2 - l2) / (r2 - l2 + 1), avg_val1 / (r2 - l2 + 1));
                } else {
                    int l = in.nextInt();
                    int r = in.nextInt();
                    l--;
                    r--;
                    out.println(new DecimalFormat("#0.0000000").format(get(1, 0, n - 1, l, r)));
                }
                q--;
            }
        }

    }

    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        public InputReader(InputStream inputStream) {
            br = new BufferedReader(new InputStreamReader(inputStream));
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

    }
}