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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
double lazy_p[4 * N];
double lazy_q[4 * N];
double ST[4 * N];
int a[N];
void Up(int x, int y, int n) { ST[n] = ST[2 * n] + ST[2 * n + 1]; }
void Set(int x, int y, int n, double p, double q) {
  ST[n] = ST[n] * p + (y - x + 1) * q;
  lazy_p[n] *= p;
  lazy_q[n] = p * lazy_q[n] + q;
}
void Down(int x, int y, int n) {
  int m = (x + y) / 2;
  Set(x, m, 2 * n, lazy_p[n], lazy_q[n]);
  Set(m + 1, y, 2 * n + 1, lazy_p[n], lazy_q[n]);
  lazy_p[n] = 1;
  lazy_q[n] = 0;
}
void init(int x, int y, int n) {
  lazy_p[n] = 1;
  if (x == y)
    ST[n] = a[x];
  else {
    init(x, (x + y) / 2, 2 * n);
    init((x + y) / 2 + 1, y, 2 * n + 1);
    Up(x, y, n);
  }
}
int qs, qe;
double p, q;
void update(int x, int y, int n) {
  if (qs > y || qe < x || qs > qe)
    return;
  else if (x >= qs && y <= qe) {
    Set(x, y, n, p, q);
  } else {
    Down(x, y, n);
    update(x, (x + y) / 2, 2 * n);
    update((x + y) / 2 + 1, y, 2 * n + 1);
    Up(x, y, n);
  }
}
double query(int x, int y, int n) {
  if (qs > y || qe < x || qs > qe)
    return 0;
  else if (x >= qs && y <= qe) {
    return ST[n];
  } else {
    Down(x, y, n);
    return query(x, (x + y) / 2, 2 * n) + query((x + y) / 2 + 1, y, 2 * n + 1);
  }
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  init(1, n, 1);
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int l1, r1, l2, r2;
      scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
      qs = l1, qe = r1;
      double sum1 = query(1, n, 1);
      qs = l2, qe = r2;
      double sum2 = query(1, n, 1);
      qs = l1, qe = r1, ::p = (double)(r1 - l1) / (r1 - l1 + 1),
      ::q = sum2 / (r2 - l2 + 1) / (r1 - l1 + 1);
      update(1, n, 1);
      qs = l2, qe = r2, ::p = (double)(r2 - l2) / (r2 - l2 + 1),
      ::q = sum1 / (r1 - l1 + 1) / (r2 - l2 + 1);
      update(1, n, 1);
    } else {
      scanf("%d %d", &qs, &qe);
      printf("%.10lf\n", query(1, n, 1));
    }
  }
}