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 maxN = 1e5 + 11;
int N, Q, A[maxN];
pair<double, pair<double, double> > IT[maxN * 4];
void update(int k, int l, int r) {
  IT[k].first =
      IT[k].first * IT[k].second.first + IT[k].second.second * (r - l + 1);
  if (l < r) {
    IT[k * 2].second.second =
        IT[k * 2].second.second * IT[k].second.first + IT[k].second.second;
    IT[k * 2].second.first *= IT[k].second.first;
    IT[k * 2 + 1].second.second =
        IT[k * 2 + 1].second.second * IT[k].second.first + IT[k].second.second;
    IT[k * 2 + 1].second.first *= IT[k].second.first;
  }
  IT[k].second = pair<double, double>(1, (1 - 1));
}
void update(int k, int l, int r, int i, int j, pair<double, double> v) {
  if (i <= l && j >= r) {
    IT[k].second.second = IT[k].second.second * v.first + v.second;
    IT[k].second.first *= v.first;
  }
  update(k, l, r);
  if (j < l || i > r) return;
  if (i <= l && j >= r) return;
  int mid = (l + r) / 2;
  update(k * 2, l, mid, i, j, v);
  update(k * 2 + 1, mid + 1, r, i, j, v);
  IT[k].first = IT[k * 2 + 1].first + IT[k * 2].first;
}
double get_it(int k, int l, int r, int i, int j) {
  update(k, l, r);
  if (j < l || i > r) return (1 - 1);
  if (i <= l && j >= r) return IT[k].first;
  int mid = (l + r) / 2;
  return get_it(k * 2, l, mid, i, j) + get_it(k * 2 + 1, mid + 1, r, i, j);
}
int main() {
  cin >> N >> Q;
  for (int i = 1; i <= N; i++)
    cin >> A[i], update(1, 1, N, i, i, pair<double, double>(1, A[i]));
  while (Q--) {
    int cmd;
    cin >> cmd;
    if (cmd == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      double x = get_it(1, 1, N, l1, r1);
      double y = get_it(1, 1, N, l2, r2);
      update(1, 1, N, l1, r1,
             pair<double, double>(double(r1 - l1) / (r1 - l1 + 1),
                                  y / (r2 - l2 + 1) / (r1 - l1 + 1)));
      update(1, 1, N, l2, r2,
             pair<double, double>(double(r2 - l2) / (r2 - l2 + 1),
                                  x / (r1 - l1 + 1) / (r2 - l2 + 1)));
    } else {
      int l, r;
      cin >> l >> r;
      printf("%.9lf\n", get_it(1, 1, N, l, r));
    }
  }
}