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>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
using namespace std;
typedef double ld;
const ld eps = 1e-7;
const int inf = 1000000010;
const long long INF = 10000000000000010LL;
const int mod = 1000000007;
const int MAXN = 100010, LOG = 20;
int n, m, k, u, v, x, y, t, a, b, l, r;
ld A[MAXN];
ld seg[MAXN << 2];
ld lazy1[MAXN << 2];
ld lazy2[MAXN << 2];
ld build(int id, int tl, int tr) {
  lazy1[id] = 1;
  if (tr - tl == 1) return seg[id] = A[tl];
  int mid = (tl + tr) >> 1;
  return seg[id] = build(id << 1, tl, mid) + build(id << 1 | 1, mid, tr);
}
void add_lazy(int id, int tl, int tr, ld Mul, ld Add = 0) {
  seg[id] *= Mul;
  lazy1[id] *= Mul;
  lazy2[id] *= Mul;
  seg[id] += Add * (tr - tl);
  lazy2[id] += Add;
}
void shift(int id, int tl, int tr) {
  int mid = (tl + tr) >> 1;
  add_lazy(id << 1, tl, mid, lazy1[id], lazy2[id]);
  add_lazy(id << 1 | 1, mid, tr, lazy1[id], lazy2[id]);
  lazy1[id] = 1;
  lazy2[id] = 0;
}
void update(int id, int tl, int tr, int l, int r, int type, ld val) {
  if (r <= tl || tr <= l) return;
  if (l <= tl && tr <= r) {
    if (type == 1) add_lazy(id, tl, tr, val, 0);
    if (type == 2) add_lazy(id, tl, tr, 1, val);
    return;
  }
  shift(id, tl, tr);
  int mid = (tl + tr) >> 1;
  update(id << 1, tl, mid, l, r, type, val);
  update(id << 1 | 1, mid, tr, l, r, type, val);
  seg[id] = seg[id << 1] + seg[id << 1 | 1];
}
ld get(int id, int tl, int tr, int l, int r) {
  if (r <= tl || tr <= l) return 0;
  if (l <= tl && tr <= r) return seg[id];
  shift(id, tl, tr);
  int mid = (tl + tr) >> 1;
  return get(id << 1, tl, mid, l, r) + get(id << 1 | 1, mid, tr, l, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> A[i];
  build(1, 1, n + 1);
  while (m--) {
    cin >> t >> l >> r;
    r++;
    if (t == 2)
      cout << setprecision(5) << fixed << get(1, 1, n + 1, l, r) << '\n';
    else {
      cin >> x >> y;
      y++;
      ld s1 = get(1, 1, n + 1, l, r), s2 = get(1, 1, n + 1, x, y);
      update(1, 1, n + 1, l, r, 1, 1 - 1. / (r - l));
      update(1, 1, n + 1, l, r, 2, s2 / (1. * (r - l) * (y - x)));
      update(1, 1, n + 1, x, y, 1, 1 - 1. / (y - x));
      update(1, 1, n + 1, x, y, 2, s1 / (1. * (r - l) * (y - x)));
    }
  }
  return 0;
}