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;
ostream& operator<<(ostream& os, vector<long long>& x) {
  for (long long i = 0; i < x.size(); i++) os << x[i] << ' ';
  return os;
}
ostream& operator<<(ostream& os, pair<long long, long long> x) {
  cout << x.first << ' ' << x.second << ' ';
  return os;
}
ostream& operator<<(ostream& os, vector<pair<long long, long long> >& x) {
  for (long long i = 0; i < x.size(); i++) os << x[i] << '\n';
  return os;
}
long long n, m;
double a[100005];
class ITrangeSum {
 private:
  double f[4 * 100005];
  pair<double, double> lazy[4 * 100005];
  bool act[4 * 100005];
  void dolazy(long long k, long long l, long long r) {
    if (!act[k]) return;
    f[k] = f[k] * lazy[k].first + lazy[k].second * (r - l + 1);
    if (l != r) {
      lazy[k * 2].first *= lazy[k].first;
      lazy[k * 2].second *= lazy[k].first;
      lazy[k * 2].second += lazy[k].second;
      lazy[k * 2 + 1].first *= lazy[k].first;
      lazy[k * 2 + 1].second *= lazy[k].first;
      lazy[k * 2 + 1].second += lazy[k].second;
      act[k * 2] = act[k * 2 + 1] = true;
    }
    lazy[k].first = 1.0;
    lazy[k].second = 0.0;
    act[k] = false;
  }

 public:
  void init(long long k, long long l, long long r) {
    lazy[k].first = 1.0;
    lazy[k].second = 0.0;
    if (l == r) {
      f[k] = a[l];
      return;
    }
    long long mid = (l + r) / 2;
    init(k * 2, l, mid);
    init(k * 2 + 1, mid + 1, r);
    f[k] = f[k * 2] + f[k * 2 + 1];
    return;
  }
  void update(long long k, long long l, long long r, long long L, long long R,
              pair<double, double> val) {
    dolazy(k, l, r);
    if (r < L || R < l) return;
    if (L <= l && r <= R) {
      f[k] = f[k] * val.first + val.second * (r - l + 1);
      if (l != r) {
        lazy[k * 2].first *= val.first;
        lazy[k * 2].second *= val.first;
        lazy[k * 2].second += val.second;
        lazy[k * 2 + 1].first *= val.first;
        lazy[k * 2 + 1].second *= val.first;
        lazy[k * 2 + 1].second += val.second;
        act[k * 2] = act[k * 2 + 1] = true;
      }
      return;
    }
    long long mid = (l + r) / 2;
    update(k * 2, l, mid, L, R, val);
    update(k * 2 + 1, mid + 1, r, L, R, val);
    f[k] = f[k * 2] + f[k * 2 + 1];
  }
  double get(long long k, long long l, long long r, long long L, long long R) {
    dolazy(k, l, r);
    if (r < L || R < l) return 0.0;
    if (L <= l && r <= R) return f[k];
    long long mid = (l + r) / 2;
    return get(k * 2, l, mid, L, R) + get(k * 2 + 1, mid + 1, r, L, R);
  }
} IT;
void treeUpdate(long long l1, long long r1, long long l2, long long r2) {
  double sumleft = 0.0, sumright = 0.0;
  sumleft = IT.get(1, 1, n, l1, r1);
  sumright = IT.get(1, 1, n, l2, r2);
  long long len1 = r1 - l1 + 1;
  long long len2 = r2 - l2 + 1;
  IT.update(1, 1, n, l1, r1,
            make_pair((len1 - 1) * 1.0 / len1, sumright / (len2 * len1)));
  IT.update(1, 1, n, l2, r2,
            make_pair((len2 - 1) * 1.0 / len2, sumleft / (len2 * len1)));
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  IT.init(1, 1, n);
  for (long long i = 1; i <= m; i++) {
    long long type;
    cin >> type;
    if (type == 1) {
      long long l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      treeUpdate(l1, r1, l2, r2);
    } else {
      long long l, r;
      cin >> l >> r;
      double ret = IT.get(1, 1, n, l, r);
      cout << setprecision(10) << fixed << ret << '\n';
    }
  }
}