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;
struct node {
  int l, r;
  long double val;
  long double lzyp, lzys;
};
int N, Q;
int arr[100005];
node seg[400005];
void pd(int idx) {
  for (int nxt = 2 * idx; nxt <= 2 * idx + 1; nxt++) {
    seg[nxt].val *= seg[idx].lzyp;
    seg[nxt].val += seg[idx].lzys * (seg[nxt].r - seg[nxt].l + 1);
    seg[nxt].lzyp *= seg[idx].lzyp;
    seg[nxt].lzys *= seg[idx].lzyp;
    seg[nxt].lzys += seg[idx].lzys;
  }
  seg[idx].lzyp = 1, seg[idx].lzys = 0;
}
void build(int l, int r, int idx) {
  seg[idx].l = l, seg[idx].r = r;
  seg[idx].lzyp = 1;
  if (l == r) {
    seg[idx].val = arr[l];
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, 2 * idx);
  build(mid + 1, r, 2 * idx + 1);
  seg[idx].val = seg[2 * idx].val + seg[2 * idx + 1].val;
}
void upd(int l, int r, long double a, long double m, int idx) {
  if (seg[idx].l == l && seg[idx].r == r) {
    seg[idx].val *= m;
    seg[idx].val += a * (seg[idx].r - seg[idx].l + 1);
    seg[idx].lzys *= m;
    seg[idx].lzyp *= m;
    seg[idx].lzys += a;
    return;
  }
  pd(idx);
  int mid = seg[idx].l + seg[idx].r >> 1;
  if (r <= mid) {
    upd(l, r, a, m, 2 * idx);
  } else if (l > mid) {
    upd(l, r, a, m, 2 * idx + 1);
  } else {
    upd(l, mid, a, m, 2 * idx);
    upd(mid + 1, r, a, m, 2 * idx + 1);
  }
  seg[idx].val = seg[2 * idx].val + seg[2 * idx + 1].val;
}
long double query(int l, int r, int idx) {
  if (seg[idx].l == l && seg[idx].r == r) {
    return seg[idx].val;
  }
  pd(idx);
  int mid = seg[idx].l + seg[idx].r >> 1;
  if (r <= mid) {
    return query(l, r, 2 * idx);
  } else if (l > mid) {
    return query(l, r, 2 * idx + 1);
  } else {
    return query(l, mid, 2 * idx) + query(mid + 1, r, 2 * idx + 1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(9);
  cin >> N >> Q;
  for (int i = 1; i <= N; i++) {
    cin >> arr[i];
  }
  build(1, N, 1);
  while (Q--) {
    int c;
    cin >> c;
    if (c == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      long double v1 = query(l1, r1, 1), v2 = query(l2, r2, 1);
      long double k1 = 1.0 * (r1 - l1) / (r1 - l1 + 1),
                  k2 = 1.0 * (r2 - l2) / (r2 - l2 + 1);
      upd(l1, r1, (1 - k1) * v2 / (r2 - l2 + 1), k1, 1);
      upd(l2, r2, (1 - k2) * v1 / (r1 - l1 + 1), k2, 1);
    } else {
      int l, r;
      cin >> l >> r;
      cout << query(l, r, 1) << "\n";
    }
  }
}