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;
typedef struct seg_node {
  double sum = 0;
  double add = 0;
  double mul = 1.0;
} sn;
sn merge(sn a, sn b) {
  sn tmp;
  tmp.sum = a.sum + b.sum;
  return tmp;
}
long long n, q;
class lazy_segment_tree {
 private:
  vector<sn> tree;
  vector<double> a;

 public:
  lazy_segment_tree(long long n) {
    tree.resize(4 * n);
    a.resize(n, 0);
  }
  lazy_segment_tree(vector<double> &v) {
    a = v;
    n = v.size();
    tree.resize(4 * v.size());
  }
  void lazy_update(long long x, long long second, long long ee) {
    if (tree[x].add == 0 && tree[x].mul == 1) return;
    tree[x].sum *= tree[x].mul;
    tree[x].sum += tree[x].add * (ee - second + 1);
    if (second != ee) {
      tree[2 * x].mul *= tree[x].mul;
      tree[2 * x].add *= tree[x].mul;
      tree[2 * x].add += tree[x].add;
      tree[2 * x + 1].mul *= tree[x].mul;
      tree[2 * x + 1].add *= tree[x].mul;
      tree[2 * x + 1].add += tree[x].add;
    }
    tree[x].add = 0;
    tree[x].mul = 1.0;
  }
  void built(long long x = 1, long long second = 0, long long ee = n - 1) {
    if (second == ee) {
      tree[x].sum = a[second];
      return;
    }
    long long m = (second + ee) / 2;
    built(2 * x, second, m);
    built(2 * x + 1, m + 1, ee);
    tree[x] = merge(tree[2 * x], tree[2 * x + 1]);
  }
  void update(long long l, long long r, double add, double mul, long long x = 1,
              long long second = 0, long long ee = n - 1) {
    lazy_update(x, second, ee);
    if (second > r || ee < l || l > r) return;
    if (second >= l && ee <= r) {
      tree[x].mul *= mul;
      tree[x].add *= mul;
      tree[x].add += add;
      lazy_update(x, second, ee);
      return;
    }
    long long m = (second + ee) / 2;
    update(l, r, add, mul, 2 * x, second, m);
    update(l, r, add, mul, 2 * x + 1, m + 1, ee);
    tree[x] = merge(tree[2 * x], tree[2 * x + 1]);
  }
  sn query(long long l, long long r, long long x = 1, long long second = 0,
           long long ee = n - 1) {
    lazy_update(x, second, ee);
    if (second > r || ee < l || l > r) return tree[0];
    if (second >= l && ee <= r) return tree[x];
    long long m = (second + ee) / 2;
    sn p = query(l, r, 2 * x, second, m);
    sn q = query(l, r, 2 * x + 1, m + 1, ee);
    sn ans = merge(p, q);
    return ans;
  }
};
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << setprecision(20) << fixed;
  cin >> n >> q;
  vector<double> a(n);
  for (auto &i : a) cin >> i;
  lazy_segment_tree lst(a);
  lst.built();
  while (q--) {
    long long type;
    cin >> type;
    if (type == 1) {
      long long l1, l2, r1, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      l1--, r1--, l2--, r2--;
      double exp1 = lst.query(l1, r1).sum;
      double exp2 = lst.query(l2, r2).sum;
      exp1 /= (r1 - l1 + 1) * (r2 - l2 + 1);
      exp2 /= (r2 - l2 + 1) * (r1 - l1 + 1);
      double m = (r1 - l1) * 1.0 / (r1 - l1 + 1);
      lst.update(l1, r1, exp2, m);
      m = (r2 - l2) * 1.0 / (r2 - l2 + 1);
      lst.update(l2, r2, exp1, m);
    } else {
      long long l, r;
      cin >> l >> r;
      l--, r--;
      double ans = lst.query(l, r).sum;
      cout << ans << '\n';
    }
  }
}