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 node {
  double sum, lazysum, lazymul;
};
node tree[4 * 112345];
node combine(node a, node b) {
  node res;
  res.sum = a.sum + b.sum;
  res.lazysum = a.lazysum + b.lazysum;
  res.lazymul = a.lazymul * b.lazymul;
  return res;
}
void build(int root, int l, int r) {
  if (l == r) {
    double x;
    cin >> x;
    tree[root].sum = x;
    tree[root].lazysum = 0.0;
    tree[root].lazymul = 1.0;
    return;
  }
  int m = (l + r) >> 1;
  build(2 * root, l, m);
  build(2 * root + 1, m + 1, r);
  tree[root] = combine(tree[2 * root], tree[2 * root + 1]);
}
void propagate(int root, int l, int r) {
  if (tree[root].lazysum == 0) return;
  tree[root].sum *= tree[root].lazymul;
  tree[root].sum += tree[root].lazysum * (r - l + 1);
  if (l != r) {
    tree[2 * root].lazysum =
        tree[2 * root].lazysum * tree[root].lazymul + tree[root].lazysum;
    tree[2 * root].lazymul *= tree[root].lazymul;
    tree[2 * root + 1].lazysum =
        tree[2 * root + 1].lazysum * tree[root].lazymul + tree[root].lazysum;
    tree[2 * root + 1].lazymul *= tree[root].lazymul;
  }
  tree[root].lazysum = 0;
  tree[root].lazymul = 1;
}
void range_update(int root, int l, int r, int a, int b, double sum,
                  double mul) {
  propagate(root, l, r);
  if (l == a && r == b) {
    tree[root].lazysum = sum;
    tree[root].lazymul = mul;
    return;
  }
  int m = (l + r) / 2;
  if (b <= m)
    range_update(2 * root, l, m, a, b, sum, mul);
  else if (m < a)
    range_update(2 * root + 1, m + 1, r, a, b, sum, mul);
  else {
    range_update(2 * root, l, m, a, m, sum, mul);
    range_update(2 * root + 1, m + 1, r, m + 1, b, sum, mul);
  }
  propagate(root, l, r);
  propagate(2 * root, l, m);
  propagate(2 * root + 1, m + 1, r);
  tree[root] = combine(tree[2 * root], tree[2 * root + 1]);
}
node query(int root, int l, int r, int a, int b) {
  propagate(root, l, r);
  if (l == a && r == b) return tree[root];
  int m = (l + r) / 2;
  if (b <= m)
    return query(2 * root, l, m, a, b);
  else if (m < a)
    return query(2 * root + 1, m + 1, r, a, b);
  else {
    node left = query(2 * root, l, m, a, m);
    node right = query(2 * root + 1, m + 1, r, m + 1, b);
    node ans = combine(left, right);
    return ans;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  build(1, 0, n - 1);
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      l1--, r1--, l2--, r2--;
      double avg1 = query(1, 0, n - 1, l1, r1).sum / (r1 - l1 + 1);
      double avg2 = query(1, 0, n - 1, l2, r2).sum / (r2 - l2 + 1);
      range_update(1, 0, n - 1, l1, r1, avg2 / (r1 - l1 + 1),
                   double(r1 - l1) / (r1 - l1 + 1));
      range_update(1, 0, n - 1, l2, r2, avg1 / (r2 - l2 + 1),
                   double(r2 - l2) / (r2 - l2 + 1));
    } else {
      int l, r;
      cin >> l >> r;
      l--, r--;
      cout << fixed << setprecision(8) << query(1, 0, n - 1, l, r).sum << endl;
    }
  }
}