895_E. Eyes Closed

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

Example input:
4 4
1 1 2 2
1 2 2 3 3
2 1 2
1 1 2 3 4
2 1 2

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

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


Example output:
3.0000000000
3.0000000000

6.0000000000
8.0400000000

23.0000000000
14.0000000000
28.0133333333
21.5733333333
55.0000000000


Tags: data structures
probabilities
Difficulty rating: 2300

Solution:
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-4;
int n, q, ty, l1, r1, l2, r2;
double d, e, d1, nn, f, p[400010], len[400010], epv[400010], s[400010];
void pd(int a, int b) {
  if (len[b] < eps) {
    len[b] = len[a];
    epv[b] = epv[a];
  } else {
    epv[b] = (len[b] * epv[a] + (len[a] - 1) * epv[b]) / (len[a] - 1 + len[b]);
    len[b] = len[a] * len[b] / (len[a] + len[b] - 1);
  }
}
double sum(int l, int r, int n, int L, int R, int A) {
  if (!p[n]) p[n] = s[r] - s[l - 1];
  int m = l + r >> 1;
  if (len[n] > eps) {
    f = ((r - l + 1) * epv[n] - p[n]) / len[n];
    p[n] += f;
    if (l < r) {
      pd(n, n * 2);
      pd(n, n * 2 + 1);
    }
    len[n] = epv[n] = 0;
  }
  if (l == L && r == R) {
    if (A == 0) {
      epv[n] = d;
      len[n] = e;
      f = ((r - l + 1) * epv[n] - p[n]) / len[n];
      for (int i = n / 2; i; i /= 2) p[i] += f;
    }
    return p[n];
  }
  double ans = 0;
  if (L <= m) ans += sum(l, m, n * 2, L, min(m, R), A);
  if (m < R) ans += sum(m + 1, r, n * 2 + 1, max(m + 1, L), R, A);
  return ans;
}
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> s[i], s[i] += s[i - 1];
  while (q--) {
    cin >> ty;
    if (ty < 2) {
      cin >> l1 >> r1 >> l2 >> r2;
      d1 = sum(1, n, 1, l1, r1, 1);
      d = sum(1, n, 1, l2, r2, 1) / (r2 - l2 + 1);
      e = r1 - l1 + 1;
      sum(1, n, 1, l1, r1, 0);
      d = d1 / (r1 - l1 + 1);
      e = r2 - l2 + 1;
      sum(1, n, 1, l2, r2, 0);
    } else {
      cin >> l1 >> r1;
      printf("%.7lf\n", sum(1, n, 1, l1, r1, 1));
    }
  }
}