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;
int n;
double seg[131072 << 1], alazy[131072 << 1], mlazy[131072 << 1];
double a[131072];
int size(int n) {
  int siz = 1;
  for (; siz < n; siz <<= 1)
    ;
  return siz << 1;
}
void build(int id, int l, double r) {
  mlazy[id] = 1;
  if (l == r - 1) {
    seg[id] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build(2 * id, l, mid);
  build(2 * id + 1, mid, r);
  seg[id] = seg[2 * id] + seg[2 * id + 1];
}
void lazy_updatea(int id, int l, int r, double x) {
  seg[id] += (r - l) * x;
  alazy[id] += x;
}
void lazy_updatem(int id, int l, int r, double x) {
  seg[id] = (seg[id]) * x;
  mlazy[id] *= x;
  alazy[id] *= x;
}
void shift(int id, int l, int r) {
  int mid = (l + r) / 2;
  lazy_updatem(2 * id, l, mid, mlazy[id]);
  lazy_updatem(2 * id + 1, mid, r, mlazy[id]);
  mlazy[id] = 1.0;
  lazy_updatea(2 * id, l, mid, alazy[id]);
  lazy_updatea(2 * id + 1, mid, r, alazy[id]);
  alazy[id] = 0;
}
void range_update_mult(int x, int y, double val, int id = 1, int l = 1,
                       int r = n + 1) {
  if (y - 1 < l || r - 1 < x) return;
  if (l >= x && r - 1 <= y - 1) {
    lazy_updatem(id, l, r, val);
    return;
  }
  shift(id, l, r);
  int mid = (l + r) / 2;
  range_update_mult(x, y, val, 2 * id, l, mid);
  range_update_mult(x, y, val, 2 * id + 1, mid, r);
  seg[id] = seg[2 * id] + seg[2 * id + 1];
}
void range_update_ad(int x, int y, double val, int id = 1, int l = 1,
                     int r = n + 1) {
  if (y - 1 < l || r - 1 < x) return;
  if (l >= x && r - 1 <= y - 1) {
    lazy_updatea(id, l, r, val);
    return;
  }
  shift(id, l, r);
  int mid = (l + r) / 2;
  range_update_ad(x, y, val, 2 * id, l, mid);
  range_update_ad(x, y, val, 2 * id + 1, mid, r);
  seg[id] = seg[2 * id] + seg[2 * id + 1];
}
double sum_lazy(int x, int y, int id = 1, int l = 1, int r = n + 1) {
  if (l >= x && r - 1 <= y - 1) return seg[id];
  if (l > y - 1 || r - 1 < x) return 0;
  shift(id, l, r);
  int mid = (l + r) / 2;
  return sum_lazy(x, y, 2 * id, l, mid) + sum_lazy(x, y, 2 * id + 1, mid, r);
}
int main() {
  int q;
  cin >> n >> q;
  for (int i = int(1); i < int(n + 1); i++) {
    cin >> a[i];
  }
  int size_ofstree = size(n);
  memset(alazy, 0, sizeof alazy);
  build(1, 1, n + 1);
  int t, l, r, ll, rr, lll, rrr;
  while (q--) {
    cin >> t;
    if (t == 1) {
      cin >> l >> r >> ll >> rr;
      double currexpsum1 = sum_lazy(l, r + 1) / (r - l + 1);
      double currexpsum2 = sum_lazy(ll, rr + 1) / (rr - ll + 1);
      range_update_mult(l, r + 1, (double)(r - l) / (double)(r - l + 1));
      range_update_mult(ll, rr + 1, (double)(rr - ll) / (double)(rr - ll + 1));
      range_update_ad(l, r + 1, currexpsum2 / (double)(r - l + 1));
      range_update_ad(ll, rr + 1, currexpsum1 / (double)(rr - ll + 1));
    } else {
      cin >> lll >> rrr;
      printf("%.9lf\n", sum_lazy(lll, rrr + 1));
    }
  }
  return 0;
}