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>
double* segtree;
std::pair<double, double>* lazy;
void build(int32_t node, int32_t left, int32_t right, int32_t* arr) {
  lazy[node] = {-1.0f, -1.0f};
  if (right - left == 1) {
    segtree[node] = arr[left];
    return;
  }
  int32_t m = (left + right) / 2;
  build(node * 2 + 1, left, m, arr);
  build(node * 2 + 2, m, right, arr);
  segtree[node] = segtree[node * 2 + 1] + segtree[node * 2 + 2];
}
void propagate(int32_t node, int32_t left, int32_t m, int32_t right) {
  if (lazy[node].first == -1.0f) return;
  segtree[node * 2 + 1] =
      segtree[node * 2 + 1] * lazy[node].first + lazy[node].second * (m - left);
  if (lazy[node * 2 + 1].first == -1.0f)
    lazy[node * 2 + 1] = lazy[node];
  else {
    lazy[node * 2 + 1].first *= lazy[node].first;
    lazy[node * 2 + 1].second =
        lazy[node * 2 + 1].second * lazy[node].first + lazy[node].second;
  }
  segtree[node * 2 + 2] = segtree[node * 2 + 2] * lazy[node].first +
                          lazy[node].second * (right - m);
  if (lazy[node * 2 + 2].first == -1.0f)
    lazy[node * 2 + 2] = lazy[node];
  else {
    lazy[node * 2 + 2].first *= lazy[node].first;
    lazy[node * 2 + 2].second =
        lazy[node * 2 + 2].second * lazy[node].first + lazy[node].second;
  }
  lazy[node] = {-1.0f, -1.0f};
}
double get_sum(int32_t node, int32_t left, int32_t right, int32_t query_left,
               int32_t query_right) {
  if (left >= query_right || right <= query_left) return 0;
  if (left >= query_left && right <= query_right) return segtree[node];
  int32_t m = (left + right) / 2;
  propagate(node, left, m, right);
  return get_sum(node * 2 + 1, left, m, query_left, query_right) +
         get_sum(node * 2 + 2, m, right, query_left, query_right);
}
void update(int32_t node, int32_t left, int32_t right, int32_t query_left,
            int32_t query_right, double mul, double add) {
  if (left >= query_right || right <= query_left) return;
  if (left >= query_left && right <= query_right) {
    segtree[node] = segtree[node] * mul + add * (right - left);
    if (lazy[node].first == -1.0f)
      lazy[node] = {mul, add};
    else {
      lazy[node].first *= mul;
      lazy[node].second = lazy[node].second * mul + add;
    }
    return;
  }
  int32_t m = (left + right) / 2;
  propagate(node, left, m, right);
  update(node * 2 + 1, left, m, query_left, query_right, mul, add);
  update(node * 2 + 2, m, right, query_left, query_right, mul, add);
  segtree[node] = segtree[node * 2 + 1] + segtree[node * 2 + 2];
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  int32_t n, num_queries;
  std::cin >> n >> num_queries;
  int32_t* arr = new int32_t[n];
  for (int32_t i = 0; i < n; i++) std::cin >> arr[i];
  segtree = new double[4 * n];
  lazy = new std::pair<double, double>[4 * n];
  build(0, 0, n, arr);
  std::cout << std::setprecision(20);
  for (int32_t q = 0; q < num_queries; q++) {
    int32_t type, left, right;
    std::cin >> type >> left >> right;
    left--;
    right--;
    if (type == 1) {
      int32_t left2, right2;
      std::cin >> left2 >> right2;
      left2--;
      right2--;
      double sum1 = get_sum(0, 0, n, left, right + 1);
      double sum2 = get_sum(0, 0, n, left2, right2 + 1);
      int32_t length1 = right - left + 1, length2 = right2 - left2 + 1;
      update(0, 0, n, left, right + 1, (length1 - 1) / (double)(length1),
             sum2 / (double)((int64_t)length1 * length2));
      update(0, 0, n, left2, right2 + 1, (length2 - 1) / (double)(length2),
             sum1 / (double)((int64_t)length1 * length2));
    } else {
      std::cout << get_sum(0, 0, n, left, right + 1) << "\n";
    }
  }
  return 0;
}