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;
template <typename T>
void printvec(vector<T>& vec) {
  for (int i = 0; i < vec.size(); i++) {
    cout << vec[i] << " ";
  }
  cout << "\n";
}
vector<double> st, lazy0, lazy1, arr;
int n, q;
void build(int si, int second, int se) {
  if (second == se) {
    st[si] = arr[second];
    return;
  }
  int mid = (second + se) / 2;
  build(2 * si, second, mid);
  build(2 * si + 1, mid + 1, se);
  st[si] = st[2 * si] + st[2 * si + 1];
}
void lazywork(int si, int second, int se) {
  if (abs(lazy1[si] - 1) > 0.0000001) {
    st[si] = (st[si] * lazy1[si]);
    if (second != se) {
      lazy1[2 * si] = (lazy1[2 * si] * lazy1[si]);
      lazy1[2 * si + 1] = (lazy1[2 * si + 1] * lazy1[si]);
      lazy0[2 * si] = (lazy0[2 * si] * lazy1[si]);
      lazy0[2 * si + 1] = (lazy0[2 * si + 1] * lazy1[si]);
    }
    lazy1[si] = 1;
  }
  if (abs(lazy0[si]) > 0.0000001) {
    double len = se - second + 1;
    st[si] = (st[si] + len * lazy0[si]);
    if (second != se) {
      lazy0[2 * si] = (lazy0[2 * si] + lazy0[si]);
      lazy0[2 * si + 1] = (lazy0[2 * si + 1] + lazy0[si]);
    }
    lazy0[si] = 0;
  }
}
void update(int si, int second, int se, int qs, int qe, double val, int type) {
  lazywork(si, second, se);
  if (second >= qs && se <= qe) {
    if (type == 0) {
      lazy0[si] = (lazy0[si] + val);
    } else {
      lazy1[si] = (lazy1[si] * val);
    }
    return;
  }
  if (se < qs || second > qe || second >= se) return;
  int mid = (second + se) / 2;
  update(2 * si, second, mid, qs, qe, val, type);
  update(2 * si + 1, mid + 1, se, qs, qe, val, type);
  lazywork(2 * si, second, mid);
  lazywork(2 * si + 1, mid + 1, se);
  st[si] = st[2 * si] + st[2 * si + 1];
}
double query(int si, int second, int se, int qs, int qe) {
  lazywork(si, second, se);
  if (second >= qs && se <= qe) return st[si];
  if (se < qs || second > qe || second >= se) return 0;
  int mid = (second + se) / 2;
  return query(2 * si, second, mid, qs, qe) +
         query(2 * si + 1, mid + 1, se, qs, qe);
}
int main() {
  std::ios::sync_with_stdio(false);
  int ttt = 1;
  while (ttt--) {
    int i;
    cin >> n >> q;
    arr.resize(n);
    st.resize(4 * n + 4);
    lazy0.resize(4 * n + 4);
    lazy1.resize(4 * n + 4);
    for (i = 0; i < n; i++) {
      cin >> arr[i];
    }
    for (i = 0; i < lazy0.size(); i++) {
      lazy0[i] = 0;
      lazy1[i] = 1;
    }
    build(1, 0, n - 1);
    for (i = 0; i < q; i++) {
      int type;
      cin >> type;
      if (type == 1) {
        int l1, r1, l2, r2;
        cin >> l1 >> r1 >> l2 >> r2;
        l1--;
        r1--;
        l2--;
        r2--;
        double sum1 = query(1, 0, n - 1, l1, r1);
        double sum = query(1, 0, n - 1, l2, r2);
        double z = (r1 - l1) * 1.00 / (r1 - l1 + 1);
        update(1, 0, n - 1, l1, r1, z, 1);
        sum /= (r2 - l2 + 1);
        sum /= (r1 - l1 + 1);
        update(1, 0, n - 1, l1, r1, sum, 0);
        z = (r2 - l2) * 1.00 / (r2 - l2 + 1);
        update(1, 0, n - 1, l2, r2, z, 1);
        sum1 /= (r2 - l2 + 1);
        sum1 /= (r1 - l1 + 1);
        update(1, 0, n - 1, l2, r2, sum1, 0);
      } else {
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        double ans = query(1, 0, n - 1, a, b);
        cout << fixed << setprecision(20) << ans << "\n";
      }
    }
  }
  return 0;
}