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;
long long const M = 2e5 + 10, M2 = 1e7 + 10, mod = 1e9 + 7, inf = 1e9 + 10;
long double seg[M * 4], lazy[M * 4], p[M * 4];
long long tmp;
void shift(long long id1, long long id2, long long len) {
  p[id2] *= p[id1];
  seg[id2] = (p[id1]) * seg[id2] + lazy[id1] * (long double)len;
  lazy[id2] = p[id1] * lazy[id2] + lazy[id1];
}
void upd(long long l, long long r, long long st, long long en, long double inc,
         long long node) {
  if (st <= l && r <= en) {
    seg[node] = ((long double)1 / (long double)(tmp)) * inc * (r - l) +
                ((long double)(tmp - 1) / (long double)(tmp)) * seg[node];
    p[node] *= (long double)(tmp - 1) / (long double)(tmp);
    lazy[node] = ((long double)1 / (long double)(tmp)) * inc +
                 ((long double)(tmp - 1) / (long double)(tmp)) * lazy[node];
    return;
  }
  if (st >= r || l >= en) return;
  long long mid = (l + r) >> 1;
  shift(node, node * 2, mid - l);
  shift(node, node * 2 + 1, r - mid);
  p[node] = 1, lazy[node] = 0;
  upd(l, mid, st, en, inc, node * 2);
  upd(mid, r, st, en, inc, node * 2 + 1);
  seg[node] = seg[node * 2] + seg[node * 2 + 1];
}
long double get(long long l, long long r, long long st, long long en,
                long long node) {
  if (st <= l && r <= en) {
    return seg[node];
  }
  if (st >= r || l >= en) return (long double)0;
  long long mid = (l + r) >> 1;
  shift(node, node * 2, mid - l);
  shift(node, node * 2 + 1, r - mid);
  p[node] = 1, lazy[node] = 0;
  return get(l, mid, st, en, node * 2) + get(mid, r, st, en, node * 2 + 1);
}
int32_t main() {
  long long n, q;
  cin >> n >> q;
  long long tmpp;
  tmp = 1;
  for (long long i = 1; i <= n; i++)
    cin >> tmpp, upd(1, n + 1, i, i + 1, (long double)tmpp, 1);
  while (q--) {
    long long ty, l, r;
    cin >> ty;
    if (ty == 1) {
      long long l1, l2, r1, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      long double inc1 =
          get(1, n + 1, l1, r1 + 1, 1) / (long double)(r1 - l1 + 1);
      long double inc2 =
          get(1, n + 1, l2, r2 + 1, 1) / (long double)(r2 - l2 + 1);
      tmp = r1 - l1 + 1;
      upd(1, n + 1, l1, r1 + 1, inc2, 1);
      tmp = r2 - l2 + 1;
      upd(1, n + 1, l2, r2 + 1, inc1, 1);
    } else {
      cin >> l >> r;
      cout << fixed << setprecision(10) << get(1, n + 1, l, r + 1, 1) << endl;
    }
  }
}