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;
int n, a[100005], q, c;
double first, second, z, t, s1, s2, l1, l2;
pair<double, double> lz[400005];
double s[400005];
void Sift(int nod, int l, int r) {
  s[nod] = lz[nod].first * s[nod] + lz[nod].second * (r - l + 1);
  if (l != r) {
    lz[2 * nod].first = lz[2 * nod].first * lz[nod].first;
    lz[2 * nod].second = lz[2 * nod].second * lz[nod].first + lz[nod].second;
    lz[2 * nod + 1].first = lz[2 * nod + 1].first * lz[nod].first;
    lz[2 * nod + 1].second =
        lz[2 * nod + 1].second * lz[nod].first + lz[nod].second;
  }
  lz[nod] = {1, 0};
}
void Upd(int nod, int l, int r, int lt, int rt, double a, double b) {
  Sift(nod, l, r);
  if (l > rt || r < lt) return;
  if (l >= lt && r <= rt) {
    lz[nod] = {a, b};
    Sift(nod, l, r);
    return;
  }
  int mid = (l + r) / 2;
  Upd(2 * nod, l, mid, lt, rt, a, b);
  Upd(2 * nod + 1, mid + 1, r, lt, rt, a, b);
  s[nod] = s[2 * nod] + s[2 * nod + 1];
}
double Qry(int nod, int l, int r, int lt, int rt) {
  Sift(nod, l, r);
  if (l > rt || r < lt) return 0;
  if (l >= lt && r <= rt) {
    return s[nod];
  }
  int mid = (l + r) / 2;
  return Qry(2 * nod, l, mid, lt, rt) + Qry(2 * nod + 1, mid + 1, r, lt, rt);
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    Upd(1, 1, n, i, i, 0, a[i]);
  }
  cout << fixed << setprecision(7);
  for (int i = 1; i <= q; i++) {
    cin >> c;
    if (c == 1) {
      cin >> first >> second >> z >> t;
      s1 = Qry(1, 1, n, first, second);
      s2 = Qry(1, 1, n, z, t);
      l1 = second - first + 1;
      l2 = t - z + 1;
      Upd(1, 1, n, first, second, (l1 - 1) / l1, s2 / l1 / l2);
      Upd(1, 1, n, z, t, (l2 - 1) / l2, s1 / l1 / l2);
    } else {
      cin >> first >> second;
      cout << Qry(1, 1, n, first, second) << '\n';
    }
  }
}