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 long long MOD = 1e9 + 7;
class node {
  node *left, *rght;
  int l, r;
  double d1 = 1, d2 = 0;
  double v;
  bool stale = false;
  double calc() {
    if (l < r) {
      left->apply(d1, d2);
      rght->apply(d1, d2);
    }
    v = stale ? left->calc() + rght->calc() : v * d1 + d2 * (r - l + 1);
    d1 = 1;
    d2 = 0;
    stale = false;
    return v;
  }
  void apply(double v1, double v2) {
    if (v1 == 1 && v2 == 0) return;
    d1 *= v1;
    d2 = d2 * v1 + v2;
  }

 public:
  double query(int ql, int qr) {
    if (ql > r || qr < l) return 0;
    calc();
    if (ql <= l && r <= qr) return v;
    return left->query(ql, qr) + rght->query(ql, qr);
  }
  void upd(int ql, int qr, double v1, double v2) {
    if (l > qr || r < ql) return;
    if (ql <= l && r <= qr) {
      apply(v1, v2);
      return;
    }
    stale = true;
    left->upd(ql, qr, v1, v2);
    rght->upd(ql, qr, v1, v2);
  }
  node(int l, int r, double *src) : l(l), r(r) {
    if (l == r) {
      v = src[l];
      return;
    }
    int mid = (l + r) / 2;
    left = new node(l, mid, src);
    rght = new node(mid + 1, r, src);
    v = left->v + rght->v;
  }
};
long long n, q;
double a[111111];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  node head(1, n, a);
  for (int i = 0; i < q; i++) {
    int t, l1, r1;
    cin >> t >> l1 >> r1;
    if (t == 2) {
      cout << head.query(l1, r1) << '\n';
    } else {
      int l2, r2;
      cin >> l2 >> r2;
      double e1 = head.query(l1, r1), e2 = head.query(l2, r2);
      head.upd(l1, r1, (r1 - l1) * 1.0 / (r1 - l1 + 1),
               e2 / (r1 - l1 + 1) / (r2 - l2 + 1));
      head.upd(l2, r2, (r2 - l2) * 1.0 / (r2 - l2 + 1),
               e1 / (r2 - l2 + 1) / (r1 - l1 + 1));
    }
  }
  cout.flush();
  return 0;
}