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, q;
double a[1000000];
double add[1000000];
double mult[1000000];
double t[1000000];
void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = a[tl];
    mult[v] = 1;
    return;
  }
  int tm = (tl + tr) / 2;
  build(v * 2, tl, tm);
  build(v * 2 + 1, tm + 1, tr);
  t[v] = t[v * 2] + t[v * 2 + 1];
  mult[v] = 1;
}
void push(int v, int tl, int tr) {
  t[v * 2] *= mult[v];
  t[v * 2 + 1] *= mult[v];
  mult[v * 2] *= mult[v];
  mult[v * 2 + 1] *= mult[v];
  add[v * 2] *= mult[v];
  add[v * 2 + 1] *= mult[v];
  mult[v] = 1;
  int tm = (tl + tr) / 2;
  t[v * 2] += (double)(tm - tl + 1) * add[v];
  t[v * 2 + 1] += (double)(tr - tm) * add[v];
  add[v * 2] += add[v];
  add[v * 2 + 1] += add[v];
  add[v] = 0;
  t[v] = t[v * 2] + t[v * 2 + 1];
}
void update(int v, int tl, int tr, int l, int r, double mult_val,
            double add_val) {
  if (l > r) return;
  if (tl == l && tr == r) {
    t[v] *= mult_val;
    t[v] += double(tr - tl + 1) * add_val;
    mult[v] *= mult_val;
    add[v] *= mult_val;
    add[v] += add_val;
    return;
  }
  push(v, tl, tr);
  int tm = (tl + tr) / 2;
  update(v * 2, tl, tm, l, min(tm, r), mult_val, add_val);
  update(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, mult_val, add_val);
  t[v] = t[v * 2] + t[v * 2 + 1];
}
double get(int v, int tl, int tr, int l, int r) {
  if (tl == l && tr == r) return t[v];
  if (l > r) return 0;
  push(v, tl, tr);
  int tm = (tl + tr) / 2;
  return get(v * 2, tl, tm, l, min(r, tm)) +
         get(v * 2 + 1, tm + 1, tr, max(tm + 1, l), r);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  build(1, 0, n - 1);
  while (q--) {
    int type, l, r;
    cin >> type >> l >> r;
    l--;
    r--;
    if (type == 1) {
      int le, ri;
      cin >> le >> ri;
      le--;
      ri--;
      double avg_val1 = get(1, 0, n - 1, l, r) / double(r - l + 1);
      double avg_val2 = get(1, 0, n - 1, le, ri) / double(ri - le + 1);
      update(1, 0, n - 1, l, r, double(r - l) / double(r - l + 1),
             avg_val2 / double(r - l + 1));
      update(1, 0, n - 1, le, ri, double(ri - le) / double(ri - le + 1),
             avg_val1 / double(ri - le + 1));
    } else
      printf("%.7f\n", get(1, 0, n - 1, l, r));
  }
}