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 maxn = 131072;
void ad(long double& b1, long double& c1, long double& b2, long double& c2) {
  long double bt = b1 * b2;
  long double ct = b2 * c1 + c2;
  b1 = bt;
  c1 = ct;
}
struct segtree {
  vector<long double> a, b, c;
  segtree() : a(2 * maxn, 0.0), b(2 * maxn, 1.0), c(2 * maxn, 0.0) {}
  void push(int x, int xl, int xr) {
    a[x] = a[x] * b[x] + c[x] * (xr - xl + 1);
    if (x < maxn) {
      ad(b[2 * x], c[2 * x], b[x], c[x]);
      ad(b[2 * x + 1], c[2 * x + 1], b[x], c[x]);
    }
    b[x] = 1;
    c[x] = 0;
  }
  void update(int l, int r, long double bb, long double cc, int x = 1,
              int xl = 0, int xr = maxn - 1) {
    if (r < xl || xr < l) {
      push(x, xl, xr);
      return;
    }
    if (l <= xl && xr <= r) {
      ad(b[x], c[x], bb, cc);
      push(x, xl, xr);
      return;
    }
    int xm = (xl + xr) >> 1;
    push(x, xl, xr);
    update(l, r, bb, cc, 2 * x, xl, xm);
    update(l, r, bb, cc, 2 * x + 1, xm + 1, xr);
    a[x] = a[2 * x] + a[2 * x + 1];
  }
  long double query(int l, int r, int x = 1, int xl = 0, int xr = maxn - 1) {
    push(x, xl, xr);
    if (r < xl || xr < l) return 0;
    if (l <= xl && xr <= r) return a[x];
    int xm = (xl + xr) >> 1;
    return query(l, r, 2 * x, xl, xm) + query(l, r, 2 * x + 1, xm + 1, xr);
  }
} drvo;
int n, q;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    drvo.a[i + maxn] = x;
  }
  for (int i = maxn - 1; i; i--) {
    drvo.a[i] = drvo.a[2 * i] + drvo.a[2 * i + 1];
  }
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int a, b, c, d;
      cin >> a >> b >> c >> d;
      double sab = drvo.query(a, b);
      double scd = drvo.query(c, d);
      long double n = b - a + 1;
      long double m = d - c + 1;
      drvo.update(a, b, (n - 1) / n, scd / m / n);
      drvo.update(c, d, (m - 1) / m, sab / n / m);
    } else {
      int a, b;
      cin >> a >> b;
      cout << drvo.query(a, b) << '\n';
    }
  }
}