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;
int a[100005];
struct oper {
  long double pro, add;
  oper(int x) : pro(1), add(0) {}
  oper(long double pro = 1, long double add = 0) : pro(pro), add(add) {}
  oper &operator+=(oper &a) {
    this->pro = this->pro * a.pro;
    this->add = this->add * a.pro + a.add;
    return *this;
  }
  oper operator=(int x) {
    if (!x) this->pro = 0, this->add = 0;
  }
};
struct node {
  node *l, *r;
  long double val;
  oper ope;
  node() : l(0), r(0), val(0), ope(0) {}
} * root;
void addop(node *now, oper &op, int wid) {
  now->val = now->val * op.pro + op.add * wid;
  now->ope += op;
}
void push(node *now, int l, int r) {
  if (now->ope.pro != 1.0 || now->ope.add != 0.0) {
    addop(now->l, now->ope, ((l + r) >> 1) - l + 1);
    addop(now->r, now->ope, r - ((l + r) >> 1));
    now->ope.pro = 1.0;
    now->ope.add = 0.0;
  }
}
void pull(node *now) { now->val = now->l->val + now->r->val; }
void build(node *now, int l, int r) {
  if (l == r) {
    now->val = a[l];
    return;
  }
  build(now->l = new node(), l, ((l + r) >> 1));
  build(now->r = new node(), ((l + r) >> 1) + 1, r);
  pull(now);
}
void modify(node *now, int l, int r, int ml, int mr, oper op) {
  if (mr < l || r < ml) return;
  if (ml <= l && r <= mr) return addop(now, op, r - l + 1), void();
  push(now, l, r);
  modify(now->l, l, ((l + r) >> 1), ml, mr, op);
  modify(now->r, ((l + r) >> 1) + 1, r, ml, mr, op);
  pull(now);
}
long double query(node *now, int l, int r, int ql, int qr) {
  if (qr < l || r < ql) return 0;
  if (ql <= l && r <= qr) return now->val;
  push(now, l, r);
  return query(now->l, l, ((l + r) >> 1), ql, qr) +
         query(now->r, ((l + r) >> 1) + 1, r, ql, qr);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  build(root = new node(), 1, n);
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      long double lsum = query(root, 1, n, l1, r1),
                  rsum = query(root, 1, n, l2, r2);
      modify(root, 1, n, l1, r1,
             oper(1. * (r1 - l1) / (r1 - l1 + 1),
                  rsum / (r2 - l2 + 1) / (r1 - l1 + 1)));
      modify(root, 1, n, l2, r2,
             oper(1. * (r2 - l2) / (r2 - l2 + 1),
                  lsum / (r1 - l1 + 1) / (r2 - l2 + 1)));
    } else {
      int l, r;
      cin >> l >> r;
      cout << fixed << setprecision(40) << query(root, 1, n, l, r) << endl;
    }
  }
}