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 MX = 1e5 + 5;
int arr[MX];
double st[MX * 4], lzadd[MX * 4], lzmul[MX * 4];
void build(int nd, int cl, int cr) {
  lzadd[nd] = 0;
  lzmul[nd] = 1.0;
  if (cl == cr) {
    st[nd] = (double)arr[cl];
    return;
  }
  build(nd * 2, cl, (cl + cr) / 2);
  build(nd * 2 + 1, (cl + cr) / 2 + 1, cr);
  st[nd] = st[nd * 2] + st[nd * 2 + 1];
}
void prop(int nd, int cl, int cr) {
  st[nd] = st[nd] * lzmul[nd] + lzadd[nd] * (cr - cl + 1);
  if (cl != cr) {
    lzmul[nd * 2] *= lzmul[nd];
    lzmul[nd * 2 + 1] *= lzmul[nd];
    lzadd[nd * 2] = lzmul[nd] * lzadd[nd * 2] + lzadd[nd];
    lzadd[nd * 2 + 1] = lzmul[nd] * lzadd[nd * 2 + 1] + lzadd[nd];
  }
  lzmul[nd] = 1.0;
  lzadd[nd] = 0;
}
void upd(int nd, int cl, int cr, int lf, int rg, double ml, double ad) {
  prop(nd, cl, cr);
  if (rg < cl || cr < lf) return;
  if (lf <= cl && cr <= rg) {
    lzmul[nd] *= ml;
    lzadd[nd] *= ml;
    lzadd[nd] += ad;
    prop(nd, cl, cr);
    return;
  }
  upd(nd * 2, cl, (cl + cr) / 2, lf, rg, ml, ad);
  upd(nd * 2 + 1, (cl + cr) / 2 + 1, cr, lf, rg, ml, ad);
  st[nd] = st[nd * 2] + st[nd * 2 + 1];
}
double que(int nd, int cl, int cr, int lf, int rg) {
  prop(nd, cl, cr);
  if (cr < lf || rg < cl) return 0;
  if (lf <= cl && cr <= rg) return st[nd];
  return que(nd * 2, cl, (cl + cr) / 2, lf, rg) +
         que(nd * 2 + 1, (cl + cr) / 2 + 1, cr, lf, rg);
}
int n, q;
void debug() {
  for (int i = 1; i <= n; i++) cout << que(1, 1, n, i, i) << " ";
  cout << "\n";
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cout << fixed << setprecision(10);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  build(1, 1, n);
  for (int tp, l1, r1, l2, r2; q--;) {
    cin >> tp >> l1 >> r1;
    if (tp == 1) {
      cin >> l2 >> r2;
      double get1 = que(1, 1, n, l1, r1);
      double get2 = que(1, 1, n, l2, r2);
      upd(1, 1, n, l1, r1, (double)(r1 - l1) / (double)(r1 - l1 + 1.0L),
          get2 / (double)((double)(r1 - l1 + 1.0) * (double)(r2 - l2 + 1.0)));
      upd(1, 1, n, l2, r2, (double)(r2 - l2) / (double)(r2 - l2 + 1.0L),
          get1 / (double)((double)(r1 - l1 + 1.0) * (double)(r2 - l2 + 1.0)));
    } else {
      cout << que(1, 1, n, l1, r1) << "\n";
    }
  }
  return 0;
}