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 N = 100 * 1000 + 5;
int n, q;
int a[N];
long double seg[N * 4], laz1[N * 4], laz2[N * 4];
inline void act(int id, int s, int e, long double mul, long double ad) {
  seg[id] *= mul;
  laz1[id] *= mul;
  laz2[id] *= mul;
  laz2[id] += ad;
  seg[id] += ad * (e - s);
}
inline void shift(int id, int s, int e) {
  int md = (s + e) / 2;
  act(id * 2, s, md, laz1[id], laz2[id]);
  act(id * 2 + 1, md, e, laz1[id], laz2[id]);
  laz2[id] = 0;
  laz1[id] = 1;
}
void update(int l, int r, long double mul, long double ad, int id = 1,
            int s = 0, int e = n) {
  if (r <= s || e <= l) {
    return;
  }
  if (l <= s && e <= r) {
    act(id, s, e, mul, ad);
    return;
  }
  shift(id, s, e);
  int md = (s + e) / 2;
  update(l, r, mul, ad, id * 2, s, md);
  update(l, r, mul, ad, id * 2 + 1, md, e);
  seg[id] = seg[id * 2] + seg[id * 2 + 1];
}
long double get(int l, int r, int id = 1, int s = 0, int e = n) {
  if (e <= l || r <= s) {
    return 0;
  }
  if (l <= s && e <= r) {
    return seg[id];
  }
  shift(id, s, e);
  int md = (s + e) / 2;
  long double ans = get(l, r, id * 2, s, md) + get(l, r, id * 2 + 1, md, e);
  seg[id] = seg[id * 2] + seg[id * 2 + 1];
  return ans;
}
int main() {
  cin >> n >> q;
  fill(laz1, laz1 + n * 4, 1);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    update(i, i + 1, 1, a[i]);
  }
  cout << fixed << setprecision(7);
  for (int i = 0; i < q; i++) {
    int qt;
    cin >> qt;
    if (qt == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      l1--, l2--;
      long double s1 = get(l1, r1), s2 = get(l2, r2);
      update(l1, r1, (long double)(r1 - l1 - 1) / (r1 - l1),
             (long double)((long double)s2 / (r1 - l1)) / (r2 - l2));
      update(l2, r2, (long double)(r2 - l2 - 1) / (r2 - l2),
             (long double)((long double)s1 / (r1 - l1)) / (r2 - l2));
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      cout << get(l, r) << '\n';
    }
  }
  return 0;
}