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 + 17, lg = 32;
int n, q;
double it[maxn << 2];
struct lazy {
  double a, b;
  lazy() : a(1), b(0) {}
  lazy(double x, double y) : a(x), b(y) {}
} lz[maxn << 2];
lazy &operator+=(lazy &a, const lazy &b) {
  a.a *= b.a;
  a.b *= b.a;
  a.b += b.b;
  return a;
}
void init(int l = 0, int r = n, int id = 1) {
  if (l + 1 == r) {
    cin >> it[id];
    return;
  }
  int mid = l + r >> 1;
  init(l, mid, id << 1);
  init(mid, r, id << 1 | 1);
  it[id] = it[id << 1] + it[id << 1 | 1];
}
void push(int id, int l, int r) {
  int mid = l + r >> 1;
  lazy &x = lz[id];
  it[id << 1] = it[id << 1] * x.a + x.b * (mid - l);
  it[id << 1 | 1] = it[id << 1 | 1] * x.a + x.b * (r - mid);
  lz[id << 1] += lz[id];
  lz[id << 1 | 1] += lz[id];
  lz[id] = lazy();
}
void upd(int s, int e, lazy x, int l = 0, int r = n, int id = 1) {
  if (s <= l && r <= e) {
    it[id] = it[id] * x.a + x.b * (r - l);
    lz[id] += x;
    return;
  }
  if (e <= l || r <= s) return;
  int mid = l + r >> 1;
  push(id, l, r);
  upd(s, e, x, l, mid, id << 1);
  upd(s, e, x, mid, r, id << 1 | 1);
  it[id] = it[id << 1] + it[id << 1 | 1];
}
double get(int s, int e, int l = 0, int r = n, int id = 1) {
  if (s <= l && r <= e) return it[id];
  if (e <= l || r <= s) return 0;
  int mid = l + r >> 1;
  push(id, l, r);
  return get(s, e, l, mid, id << 1) + get(s, e, mid, r, id << 1 | 1);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cout << fixed << setprecision(5);
  cin >> n >> q;
  init();
  while (q--) {
    int t, l, r, x, y;
    cin >> t >> l >> r;
    l--;
    if (t == 1) {
      cin >> x >> y;
      x--;
      double a = get(l, r) / (r - l), b = get(x, y) / (y - x);
      upd(l, r, lazy({(double)(r - l - 1) / (r - l), b / (r - l)}));
      upd(x, y, lazy({(double)(y - x - 1) / (y - x), a / (y - x)}));
    } else
      cout << get(l, r) << '\n';
  }
}