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 Vertex {
  long double summ;
  long double p1;
  long double p2;
  bool p;
};
vector<Vertex> rmq;
vector<long long> data;
void create(long long i, long long l, long long r) {
  if (r - l == 1) {
    rmq[i].summ = data[l];
  } else {
    long long mid = (l + r) / 2;
    create(2 * i + 1, l, mid);
    create(2 * i + 2, mid, r);
    rmq[i].summ = rmq[2 * i + 1].summ + rmq[2 * i + 2].summ;
  }
}
void push(long long i, long long l, long long r) {
  if (!rmq[i].p) return;
  rmq[i].p = false;
  rmq[2 * i + 1].p = true;
  rmq[2 * i + 2].p = true;
  long long mid = (l + r) / 2;
  long double L1 = (mid - l);
  long double L2 = r - mid;
  rmq[2 * i + 1].summ = (rmq[i].p1 * rmq[2 * i + 1].summ) + rmq[i].p2 * L1;
  rmq[2 * i + 2].summ = (rmq[i].p1 * rmq[2 * i + 2].summ) + rmq[i].p2 * L2;
  rmq[2 * i + 1].p1 *= rmq[i].p1;
  rmq[2 * i + 2].p1 *= rmq[i].p1;
  rmq[2 * i + 1].p2 = rmq[2 * i + 1].p2 * rmq[i].p1 + rmq[i].p2;
  rmq[2 * i + 2].p2 = rmq[2 * i + 2].p2 * rmq[i].p1 + rmq[i].p2;
  rmq[i].p1 = 1;
  rmq[i].p2 = 0;
}
long double get(long long i, long long l, long long r, long long l1,
                long long r1) {
  if (r - l > 1) push(i, l, r);
  if (l1 >= r1) return 0;
  if (l1 == l && r1 == r) return rmq[i].summ;
  long long mid = (l + r) / 2;
  return get(2 * i + 1, l, mid, l1, min(r1, mid)) +
         get(2 * i + 2, mid, r, max(l1, mid), r1);
}
void prot(long long i, long long l, long long r, long long l1, long long r1,
          long double a, long double b) {
  if (r - l > 1) push(i, l, r);
  if (l1 >= r1) return;
  if (l1 == l && r1 == r) {
    rmq[i].summ = rmq[i].summ * a + b * (long double)(r - l);
    rmq[i].p = true;
    rmq[i].p1 = a;
    rmq[i].p2 = b;
    return;
  }
  long long mid = (l + r) / 2;
  prot(2 * i + 1, l, mid, l1, min(r1, mid), a, b);
  prot(2 * i + 2, mid, r, max(l1, mid), r1, a, b);
  rmq[i].summ = rmq[2 * i + 1].summ + rmq[2 * i + 2].summ;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, ai, q;
  cin >> n >> q;
  for (long long i = 0; i < n; i++) {
    cin >> ai;
    data.push_back(ai);
  }
  for (long long i = 0; i < 4 * n; i++) {
    Vertex v = {0, 1, 0, false};
    rmq.push_back(v);
  }
  create(0, 0, n);
  cout.precision(20);
  for (long long i = 0; i < q; i++) {
    long long type, lf, rf, ls, rs;
    cin >> type;
    if (type == 1) {
      cin >> lf >> rf >> ls >> rs;
      long double s1 = get(0, 0, n, lf - 1, rf) / (long double)(rf - lf + 1);
      long double s2 = get(0, 0, n, ls - 1, rs) / (long double)(rs - ls + 1);
      long double ot1 = (long double)1 / (long double)(rf - lf + 1);
      long double ot2 = (long double)1 / (long double)(rs - ls + 1);
      long double ost1 = (long double)1 - ot1;
      long double ost2 = (long double)1 - ot2;
      prot(0, 0, n, lf - 1, rf, ost1, ot1 * s2);
      prot(0, 0, n, ls - 1, rs, ost2, ot2 * s1);
    } else {
      cin >> lf >> rf;
      cout << get(0, 0, n, lf - 1, rf) << "\n";
    }
  }
  return 0;
}