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;
long long int n, q;
long double a[100005];
long double seg[1000005], pen[1000005], mul[1000005];
void build(long long int idx, long long int lo, long long int hi) {
  if (lo == hi) {
    seg[idx] = a[lo];
    pen[idx] = 0.0;
    mul[idx] = 1.0;
    return;
  }
  long long int mid = (lo + hi) / 2;
  build(2 * idx, lo, mid);
  build(2 * idx + 1, mid + 1, hi);
  seg[idx] = (seg[2 * idx] + seg[2 * idx + 1]);
  pen[idx] = 0.0;
  mul[idx] = 1.0;
}
void lazyprop(long long int idx, long long int lo, long long int hi) {
  seg[idx] *= mul[idx];
  seg[idx] += pen[idx] * ((long double)(hi - lo + 1.0));
  if (lo == hi) {
    pen[idx] = 0.0;
    mul[idx] = 1.0;
    return;
  }
  pen[2 * idx] *= mul[idx];
  pen[2 * idx + 1] *= mul[idx];
  pen[2 * idx] += pen[idx];
  pen[2 * idx + 1] += pen[idx];
  mul[2 * idx] *= mul[idx];
  mul[2 * idx + 1] *= mul[idx];
  pen[idx] = 0.0;
  mul[idx] = 1.0;
}
void update(long long int idx, long long int lo, long long int hi,
            long long int st, long long int en, long double vm,
            long double va) {
  if (lo == st && hi == en) {
    if (lo == hi) {
      seg[idx] *= mul[idx];
      seg[idx] += pen[idx];
      seg[idx] *= vm;
      seg[idx] += va;
      mul[idx] = 1.0;
      pen[idx] = 0.0;
    } else {
      lazyprop(idx, lo, hi);
      mul[idx] = vm;
      pen[idx] = va;
    }
    return;
  }
  lazyprop(idx, lo, hi);
  long long int mid = (lo + hi) / 2;
  if (en <= mid)
    update(2 * idx, lo, mid, st, en, vm, va);
  else if (st > mid)
    update(2 * idx + 1, mid + 1, hi, st, en, vm, va);
  else {
    update(2 * idx, lo, mid, st, mid, vm, va);
    update(2 * idx + 1, mid + 1, hi, mid + 1, en, vm, va);
  }
  seg[idx] = seg[2 * idx] * mul[2 * idx] + pen[2 * idx] * (mid - lo + 1.0) +
             seg[2 * idx + 1] * mul[2 * idx + 1] +
             pen[2 * idx + 1] * (long double)(hi - mid);
}
long double query(long long int idx, long long int lo, long long int hi,
                  long long int st, long long int en) {
  if (lo == st && hi == en)
    return seg[idx] * mul[idx] + pen[idx] * (hi - lo + 1.0);
  lazyprop(idx, lo, hi);
  long long int mid = (lo + hi) / 2;
  if (en <= mid)
    return query(2 * idx, lo, mid, st, en);
  else if (st > mid)
    return query(2 * idx + 1, mid + 1, hi, st, en);
  else
    return (query(2 * idx, lo, mid, st, mid) +
            query(2 * idx + 1, mid + 1, hi, mid + 1, en));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << setiosflags(ios::fixed | ios::showpoint) << setprecision(11);
  cin >> n >> q;
  for (long long int i = 1; i <= n; ++i) {
    long long int x;
    cin >> x;
    a[i] = (long double)x;
  }
  build(1, 1, n);
  while (q--) {
    long long int ty;
    cin >> ty;
    if (ty == 1) {
      long long int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      long double sz1 = (long double)r1 - l1 + 1,
                  sz2 = (long double)r2 - l2 + 1;
      long double av1 = query(1, 1, n, l1, r1) / (sz1);
      long double av2 = query(1, 1, n, l2, r2) / (sz2);
      update(1, 1, n, l1, r1, (sz1 - 1.0) / sz1, (1.0 * av2) / sz1);
      update(1, 1, n, l2, r2, (sz2 - 1.0) / sz2, (1.0 * av1) / sz2);
    } else {
      long long int l, r;
      cin >> l >> r;
      cout << query(1, 1, n, l, r) << '\n';
    }
  }
}