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 long long MAXN = 200005;
using ld = long double;
long long n, q;
long long a[MAXN];
ld IT[MAXN * 4], lazyadd[MAXN * 4], lazymul[MAXN * 4];
void build(long long l, long long r, long long id) {
  if (l == r) {
    IT[id] = a[l];
    lazymul[id] = 1;
    lazyadd[id] = 0;
    return;
  }
  long long mid = l + r >> 1;
  build(l, mid, id << 1);
  build(mid + 1, r, id << 1 | 1);
  IT[id] = IT[id << 1] + IT[id << 1 | 1];
  lazymul[id] = 1;
  lazyadd[id] = 0;
}
void push(long long id, long long l, long long r) {
  IT[id] = IT[id] * lazymul[id] + lazyadd[id] * (r - l + 1);
  if (l != r) {
    lazymul[id * 2] *= lazymul[id];
    lazyadd[id * 2] = lazyadd[id * 2] * lazymul[id] + lazyadd[id];
    lazymul[id * 2 + 1] *= lazymul[id];
    lazyadd[id * 2 + 1] = lazyadd[id * 2 + 1] * lazymul[id] + lazyadd[id];
  }
  lazymul[id] = 1;
  lazyadd[id] = 0;
}
void upd(long long u, long long v, ld mul, ld add, long long id = 1,
         long long l = 1, long long r = n) {
  push(id, l, r);
  if (r < u || l > v) return;
  if (l >= u && r <= v) {
    IT[id] = IT[id] * mul + add * (r - l + 1);
    if (l != r) {
      lazymul[id * 2] *= mul;
      lazyadd[id * 2] = lazyadd[id * 2] * mul + add;
      lazymul[id * 2 + 1] *= mul;
      lazyadd[id * 2 + 1] = lazyadd[id * 2 + 1] * mul + add;
    }
    return;
  }
  long long mid = l + r >> 1;
  upd(u, v, mul, add, id << 1, l, mid);
  upd(u, v, mul, add, id << 1 | 1, mid + 1, r);
  IT[id] = IT[id << 1] + IT[id << 1 | 1];
}
ld get_sum(long long u, long long v, long long id = 1, long long l = 1,
           long long r = n) {
  push(id, l, r);
  if (r < u || l > v) return 0;
  if (l >= u && r <= v) return IT[id];
  long long mid = l + r >> 1;
  return get_sum(u, v, id << 1, l, mid) +
         get_sum(u, v, id << 1 | 1, mid + 1, r);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  build(1, n, 1);
  while (q--) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      ld s1 = get_sum(l1, r1) / (ld)(r1 - l1 + 1),
         s2 = get_sum(l2, r2) / (ld)(r2 - l2 + 1);
      upd(l1, r1, (ld)(r1 - l1) / (ld)(r1 - l1 + 1),
          (ld)s2 / (ld)(r1 - l1 + 1));
      upd(l2, r2, (ld)(r2 - l2) / (ld)(r2 - l2 + 1),
          (ld)s1 / (ld)(r2 - l2 + 1));
    } else {
      long long l, r;
      cin >> l >> r;
      cout << setprecision(10) << fixed << get_sum(l, r) << '\n';
    }
  }
}