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 N = 1e5 + 5;
int n, q, a[N];
struct st {
  double t[4 * N], lazy[4 * N], lazy2[4 * N];
  void build_tree(int l, int r, int node) {
    if (l == r) {
      lazy[node] = 1;
      lazy2[node] = 0;
      t[node] = a[l];
      return;
    }
    int mid = (l + r) / 2;
    build_tree(l, mid, node * 2);
    build_tree(mid + 1, r, node * 2 + 1);
    t[node] = t[node * 2] + t[node * 2 + 1];
    lazy[node] = 1;
    lazy2[node] = 0;
  }
  void updlazy(int l, int r, int node) {
    if (l != r) {
      if (lazy[node] != 1 || lazy2[node] != 0) {
        int mid = (l + r) >> 1;
        t[node * 2] =
            t[node * 2] * lazy[node] + (double)(mid - l + 1) * lazy2[node];
        t[node * 2 + 1] =
            t[node * 2 + 1] * lazy[node] + (double)(r - mid) * lazy2[node];
        lazy[node * 2] *= lazy[node];
        lazy[node * 2 + 1] *= lazy[node];
        lazy2[node * 2] = lazy2[node * 2] * lazy[node] + lazy2[node];
        lazy2[node * 2 + 1] = lazy2[node * 2 + 1] * lazy[node] + lazy2[node];
        lazy[node] = 1;
        lazy2[node] = 0;
      }
    }
  }
  void upd(int l, int r, int s, int e, int node, double mul, double plus) {
    updlazy(l, r, node);
    if (l > r || l > e || r < s) return;
    if (s <= l && r <= e) {
      lazy[node] *= mul;
      lazy2[node] = lazy2[node] * mul + plus;
      t[node] = t[node] * mul + double(r - l + 1) * plus;
      return;
    }
    int mid = (l + r) / 2;
    upd(l, mid, s, e, node * 2, mul, plus);
    upd(mid + 1, r, s, e, node * 2 + 1, mul, plus);
    t[node] = t[node * 2] + t[node * 2 + 1];
  }
  double get(int l, int r, int s, int e, int node) {
    updlazy(l, r, node);
    if (l > r || l > e || r < s) return 0;
    if (s <= l && r <= e) return t[node];
    int mid = (l + r) / 2;
    return get(l, mid, s, e, node * 2) + get(mid + 1, r, s, e, node * 2 + 1);
  }
} st;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(6);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  st.build_tree(0, n - 1, 1);
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      l1--;
      r1--;
      l2--;
      r2--;
      double mul = (double)(r1 - l1) / (double)(r1 - l1 + 1);
      double add = st.get(0, n - 1, l2, r2, 1) / (double)(r2 - l2 + 1);
      add /= (double)(r1 - l1 + 1);
      double mul2 = (double)(r2 - l2) / (double)(r2 - l2 + 1);
      double add2 = st.get(0, n - 1, l1, r1, 1) / (double)(r1 - l1 + 1);
      add2 /= (double)(r2 - l2 + 1);
      st.upd(0, n - 1, l1, r1, 1, mul, add);
      st.upd(0, n - 1, l2, r2, 1, mul2, add2);
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << st.get(0, n - 1, l, r, 1) << "\n";
    }
  }
  return 0;
}