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 tope = 1 << 18;
const int primero = 1 << 17;
double suma[tope];
double anyade[tope];
double multiplica[tope];
int posle[tope];
int posri[tope];
double getvalue(int pos) {
  return suma[pos] * multiplica[pos] +
         anyade[pos] * (posri[pos] - posle[pos] + 1);
}
void opera(double &mul, double &su, double newmul, double newsu) {
  mul *= newmul;
  su = (su * newmul + newsu);
}
void propaga(int pos) {
  if (primero <= pos) return;
  opera(multiplica[2 * pos], anyade[2 * pos], multiplica[pos], anyade[pos]);
  opera(multiplica[2 * pos + 1], anyade[2 * pos + 1], multiplica[pos],
        anyade[pos]);
  multiplica[pos] = 1;
  anyade[pos] = 0;
  suma[pos] = getvalue(2 * pos) + getvalue(2 * pos + 1);
}
void anyadeentre(int pos, int pos0, int pos1, double su, double pro) {
  if (posri[pos] < pos0 or pos1 < posle[pos]) return;
  if (pos0 <= posle[pos] and posri[pos] <= pos1) {
    opera(multiplica[pos], anyade[pos], pro, su);
    return;
  }
  propaga(pos);
  anyadeentre(2 * pos, pos0, pos1, su, pro);
  anyadeentre(2 * pos + 1, pos0, pos1, su, pro);
  suma[pos] = getvalue(2 * pos) + getvalue(2 * pos + 1);
}
double computaentre(int pos, int pos0, int pos1) {
  if (posri[pos] < pos0 or pos1 < posle[pos]) return 0;
  if (pos0 <= posle[pos] and posri[pos] <= pos1) return getvalue(pos);
  propaga(pos);
  return computaentre(2 * pos, pos0, pos1) +
         computaentre(2 * pos + 1, pos0, pos1);
}
const int limite = 1000000;
int n, casos;
double a[limite];
double getavg(int l, int r) { return computaentre(1, l, r) / (r - l + 1); }
void add(int l, int r, double x) { anyadeentre(1, l, r, x, 1); }
void multiply(int l, int r, double x) { anyadeentre(1, l, r, 0, x); }
int main() {
  ios::sync_with_stdio(false);
  cout.setf(ios::fixed);
  cout.precision(12);
  for (int i = tope - 1; i >= 1; i--) {
    if (primero <= i)
      posle[i] = posri[i] = i - primero;
    else {
      posle[i] = posle[2 * i];
      posri[i] = posri[2 * i + 1];
    }
  }
  cin >> n >> casos;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    anyadeentre(1, i, i, a[i], 1);
  }
  for (int cas = 0; cas < casos; cas++) {
    int tipo;
    cin >> tipo;
    if (tipo == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      double s1 = r1 - l1 + 1;
      double s2 = r2 - l2 + 1;
      double avg1 = getavg(l1, r1);
      double avg2 = getavg(l2, r2);
      multiply(l1, r1, (s1 - 1) / s1);
      add(l1, r1, avg2 / s1);
      multiply(l2, r2, (s2 - 1) / s2);
      add(l2, r2, avg1 / s2);
    } else {
      int l, r;
      cin >> l >> r;
      double s = r - l + 1;
      double avg = getavg(l, r);
      cout << s * avg << endl;
    }
  }
}