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 segtree {
  int n;
  vector<long double> sum;
  vector<long double> lz_mul;
  vector<long double> lz_sum;
  segtree(int _n) {
    n = _n;
    sum.assign(4 * n + 5, 0.0);
    lz_sum.assign(4 * n + 5, 0.0);
    lz_mul.assign(4 * n + 5, 1.0);
  }
  void push(int id, int l, int r) {
    for (int i = id << 1; i <= (id << 1 | 1); i++) {
      sum[i] *= lz_mul[id];
      lz_mul[i] *= lz_mul[id];
      lz_sum[i] *= lz_mul[id];
    }
    lz_mul[id] = 1.0;
    int mid = (l + r) >> 1;
    sum[id << 1] += (long double)(mid - l + 1) * lz_sum[id];
    sum[id << 1 | 1] += (long double)(r - mid) * lz_sum[id];
    lz_sum[id << 1] += lz_sum[id];
    lz_sum[id << 1 | 1] += lz_sum[id];
    lz_sum[id] = 0;
    sum[id] = sum[id << 1] + sum[id << 1 | 1];
  }
  void update(int id, int l, int r, int u, int v, long double mult_val,
              long double sum_val) {
    if (l > v || r < u) return;
    if (l >= u && r <= v) {
      sum[id] *= mult_val;
      lz_sum[id] *= mult_val;
      lz_mul[id] *= mult_val;
      sum[id] += (r - l + 1) * sum_val;
      lz_sum[id] += sum_val;
      return;
    }
    push(id, l, r);
    int mid = (l + r) >> 1;
    update(id << 1, l, mid, u, v, mult_val, sum_val);
    update(id << 1 | 1, mid + 1, r, u, v, mult_val, sum_val);
    sum[id] = sum[id << 1] + sum[id << 1 | 1];
  }
  void update(int l, int r, long double mult_val, long double sum_val) {
    update(1, 1, n, l, r, mult_val, sum_val);
  }
  long double getsum(int id, int l, int r, int u, int v) {
    if (l > v || r < u) return 0;
    if (l >= u && r <= v) return sum[id];
    push(id, l, r);
    int mid = (l + r) >> 1;
    long double lnode = getsum(id << 1, l, mid, u, v);
    long double rnode = getsum(id << 1 | 1, mid + 1, r, u, v);
    return lnode + rnode;
  }
  long double getsum(int l, int r) { return getsum(1, 1, n, l, r); }
};
const int N = 1e5 + 5;
int n, q;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  segtree st(n);
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    st.update(i, i, 1, x);
  }
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r, u, v;
      cin >> l >> r >> u >> v;
      long double suml = st.getsum(l, r), sumr = st.getsum(u, v);
      long double dv = (long double)(r - l + 1) * (v - u + 1);
      st.update(l, r, (long double)(r - l) / (long double)(r - l + 1),
                sumr / dv);
      st.update(u, v, (long double)(v - u) / (long double)(v - u + 1),
                suml / dv);
    } else {
      int l, r;
      cin >> l >> r;
      cout << fixed << setprecision(6) << st.getsum(l, r) << "\n";
    }
  }
  return 0;
}