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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 * @author Don Li
 */
public class EyesClosed {
    
    int n, h;
    double[] t, dm, dp;
    int[] sz;
    
    void solve() {
        n = in.nextInt(); int Q = in.nextInt();
        
        h = 32 - Integer.numberOfLeadingZeros(n);
        t = new double[n << 1];
        dm = new double[n];
        dp = new double[n];
        sz = new int[n << 1];
        for (int i = 0; i < n; i++) {
            t[n + i] = in.nextInt();
            sz[n + i] = 1;
        }
        for (int i = n - 1; i > 0; i--) {
            t[i] = t[i << 1] + t[i << 1 | 1];
            sz[i] = sz[i << 1] + sz[i << 1 | 1];
        }
        Arrays.fill(dm, 1);
        
        while (Q-- > 0) {
            int type = in.nextInt();
            if (type == 1) {
                int l1 = in.nextInt() - 1, r1 = in.nextInt(), l2 = in.nextInt() - 1, r2 = in.nextInt();
                double avg1 = sum(l1, r1) / (r1 - l1), avg2 = sum(l2, r2) / (r2 - l2);
                mul(l1, r1, (double) (r1 - l1 - 1) / (r1 - l1));
                add(l1, r1, avg2 / (r1 - l1));
                mul(l2, r2, (double) (r2 - l2 - 1) / (r2 - l2));
                add(l2, r2, avg1 / (r2 - l2));
            } else {
                int l = in.nextInt() - 1, r = in.nextInt();
                double ans = sum(l, r);
                out.println(ans);
            }
        }
    }
    
    double sum(int l, int r) {
        push(l + n);
        push(r + n - 1);
        double res = 0;
        for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
            if ((l & 1) > 0) res += t[l++];
            if ((r & 1) > 0) res += t[--r];
        }
        return res;
    }
    
    void add(int l, int r, double v) {
        int l0 = l + n, r0 = r + n - 1;
        push(l0);
        push(r0);
        int k = 1;
        for (l += n, r += n; l < r; l >>= 1, r >>= 1, k <<= 1) {
            if ((l & 1) > 0) apply_add(l++, v);
            if ((r & 1) > 0) apply_add(--r, v);
        }
        lift(l0);
        lift(r0);
    }
    
    void mul(int l, int r, double v) {
        int l0 = l + n, r0 = r + n - 1;
        push(l0);
        push(r0);
        for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
            if ((l & 1) > 0) apply_mul(l++, v);
            if ((r & 1) > 0) apply_mul(--r, v);
        }
        lift(l0);
        lift(r0);
    }
    
    void apply_mul(int i, double v) {
        t[i] *= v;
        if (i < n) {
            dm[i] *= v;
            dp[i] *= v;
        }
    }
    
    void apply_add(int i, double v) {
        t[i] += v * sz[i];
        if (i < n) {
            dp[i] += v;
        }
    }
    
    void push(int i) {
        for (int s = h; s > 0; s--) {
            int p = i >> s;
            if (dm[p] != 1) {
                apply_mul(p << 1, dm[p]);
                apply_mul(p << 1 | 1, dm[p]);
                dm[p] = 1;
            }
            if (dp[p] != 0) {
                apply_add(p << 1, dp[p]);
                apply_add(p << 1 | 1, dp[p]);
                dp[p] = 0;
            }
        }
    }
    
    void lift(int i) {
        for (int p = i >> 1; p > 0; p >>= 1) {
            t[p] = (t[p << 1] + t[p << 1 | 1]) * dm[p] + sz[p] * dp[p];
        }
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new EyesClosed().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}