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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long power(long long a, long long n) {
  a %= 1000000007;
  if (n == 1) return a;
  if (n == 0) return 1;
  if (n % 2)
    return (a * (power((a * a) % 1000000007, n / 2) % 1000000007)) % 1000000007;
  return power((a * a) % 1000000007, n / 2) % 1000000007;
}
const long long inf = (long long)1e18;
long long inverse(long long x) { return power(x, 1000000007 - 2) % 1000000007; }
long long n, q;
vector<double> v(200005);
double tree[4 * 200005];
pair<double, double> lazy[4 * 200005];
void build(long long no, long long st, long long la) {
  if (st == la) {
    tree[no] = v[st];
    lazy[no] = {1, 0};
    return;
  }
  long long mid = (st + la) / 2;
  build(2 * no, st, mid);
  build(2 * no + 1, mid + 1, la);
  tree[no] = tree[2 * no] + tree[2 * no + 1];
  lazy[no] = {1, 0};
}
void update(long long no, long long st, long long la, long long l, long long r,
            double prob, double extra) {
  tree[no] = lazy[no].first * tree[no] + lazy[no].second * (la - st + 1);
  if (st != la) {
    lazy[2 * no].first *= lazy[no].first,
        lazy[2 * no].second =
            lazy[2 * no].second * lazy[no].first + lazy[no].second;
    lazy[2 * no + 1].first *= lazy[no].first,
        lazy[2 * no + 1].second =
            lazy[2 * no + 1].second * lazy[no].first + lazy[no].second;
  }
  lazy[no] = {1, 0};
  if (l > la || r < st) return;
  if (l <= st && r >= la) {
    tree[no] = tree[no] * prob + extra * (la - st + 1);
    if (st != la) {
      lazy[2 * no].first *= prob,
          lazy[2 * no].second = lazy[2 * no].second * prob + extra;
      lazy[2 * no + 1].first *= prob,
          lazy[2 * no + 1].second = lazy[2 * no + 1].second * prob + extra;
    }
    return;
  }
  long long mid = (st + la) / 2;
  update(2 * no, st, mid, l, r, prob, extra);
  update(2 * no + 1, mid + 1, la, l, r, prob, extra);
  tree[no] = tree[2 * no] + tree[2 * no + 1];
}
double query(long long no, long long st, long long la, long long l,
             long long r) {
  tree[no] = lazy[no].first * tree[no] + lazy[no].second * (la - st + 1);
  if (st != la) {
    lazy[2 * no].first *= lazy[no].first,
        lazy[2 * no].second =
            lazy[2 * no].second * lazy[no].first + lazy[no].second;
    lazy[2 * no + 1].first *= lazy[no].first,
        lazy[2 * no + 1].second =
            lazy[2 * no + 1].second * lazy[no].first + lazy[no].second;
  }
  lazy[no] = {1, 0};
  if (l > la || r < st) return 0;
  if (l <= st && r >= la) return tree[no];
  long long mid = (st + la) / 2;
  return query(2 * no, st, mid, l, r) + query(2 * no + 1, mid + 1, la, l, r);
}
void solve() {
  cin >> n >> q;
  for (long long i = 1; i < n + 1; ++i) cin >> v[i];
  build(1, 1, n);
  while (q--) {
    long long type;
    cin >> type;
    if (type == 1) {
      double l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      double prob1 = (r2 - l2) / (r2 - l2 + 1),
             extra1 = (query(1, 1, n, l1, r1) / (r1 - l1 + 1)) / (r2 - l2 + 1);
      double prob2 = (r1 - l1) / (r1 - l1 + 1),
             extra2 = (query(1, 1, n, l2, r2) / (r2 - l2 + 1)) / (r1 - l1 + 1);
      update(1, 1, n, l2, r2, prob1, extra1);
      update(1, 1, n, l1, r1, prob2, extra2);
    } else {
      long long l, r;
      cin >> l >> r;
      cout << fixed << setprecision(10) << query(1, 1, n, l, r) << "\n";
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long T = 1;
  for (long long t = 1; t < T + 1; ++t) {
    solve();
  }
  return 0;
}