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;
template <class T, class T2>
inline void chkmax(T &x, const T2 &y) {
  if (x < y) x = y;
}
template <class T, class T2>
inline void chkmin(T &x, const T2 &y) {
  if (x > y) x = y;
}
const int MAXN = (1 << 20);
int n, q;
double a[MAXN];
struct node {
  double sum, lazy_mul, lazy_add;
  node() {
    sum = 0;
    lazy_add = 0;
    lazy_mul = 1;
  }
  node(int val) {
    sum = val;
    lazy_add = 0;
    lazy_mul = 1;
  }
};
node temp, broken;
node merge(node l, node r) {
  temp.sum = l.sum + r.sum;
  temp.lazy_add = 0;
  temp.lazy_mul = 1;
  return temp;
}
struct segment_tree {
  node tr[4 * MAXN];
  void push(int l, int r, int idx) {
    if (tr[idx].lazy_mul != 1) {
      tr[idx].sum *= tr[idx].lazy_mul;
      if (l != r) {
        tr[2 * idx + 1].lazy_mul *= tr[idx].lazy_mul;
        tr[2 * idx + 2].lazy_mul *= tr[idx].lazy_mul;
        tr[2 * idx + 1].lazy_add *= tr[idx].lazy_mul;
        tr[2 * idx + 2].lazy_add *= tr[idx].lazy_mul;
      }
      tr[idx].lazy_mul = 1;
    }
    if (tr[idx].lazy_add != 0) {
      tr[idx].sum += (double)(r - l + 1) * 1.0 * tr[idx].lazy_add;
      if (l != r) {
        tr[2 * idx + 1].lazy_add += tr[idx].lazy_add;
        tr[2 * idx + 2].lazy_add += tr[idx].lazy_add;
      }
      tr[idx].lazy_add = 0;
    }
  }
  void init(int l, int r, int idx) {
    if (l == r) {
      tr[idx] = node(a[l]);
      return;
    }
    int mid = (l + r) >> 1;
    init(l, mid, 2 * idx + 1);
    init(mid + 1, r, 2 * idx + 2);
    tr[idx] = merge(tr[2 * idx + 1], tr[2 * idx + 2]);
  }
  void add(int qL, int qR, double val, int l, int r, int idx) {
    push(l, r, idx);
    if (qL > r || l > qR) return;
    if (qL <= l && r <= qR) {
      tr[idx].lazy_add += val;
      push(l, r, idx);
      return;
    }
    int mid = (l + r) >> 1;
    add(qL, qR, val, l, mid, 2 * idx + 1);
    add(qL, qR, val, mid + 1, r, 2 * idx + 2);
    tr[idx] = merge(tr[2 * idx + 1], tr[2 * idx + 2]);
  }
  void mult(int qL, int qR, double val, int l, int r, int idx) {
    push(l, r, idx);
    if (qL > r || l > qR) return;
    if (qL <= l && r <= qR) {
      tr[idx].lazy_mul *= val;
      push(l, r, idx);
      return;
    }
    int mid = (l + r) >> 1;
    mult(qL, qR, val, l, mid, 2 * idx + 1);
    mult(qL, qR, val, mid + 1, r, 2 * idx + 2);
    tr[idx] = merge(tr[2 * idx + 1], tr[2 * idx + 2]);
  }
  node query(int qL, int qR, int l, int r, int idx) {
    push(l, r, idx);
    if (l > qR || r < qL) return broken;
    if (qL <= l && r <= qR) return tr[idx];
    int mid = (l + r) >> 1;
    return merge(query(qL, qR, l, mid, 2 * idx + 1),
                 query(qL, qR, mid + 1, r, 2 * idx + 2));
  }
};
void read() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
}
segment_tree t;
void solve() {
  t.init(1, n, 0);
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      double len1 = r1 - l1 + 1, len2 = r2 - l2 + 1;
      double Exp1 = t.query(l1, r1, 1, n, 0).sum * (1.0 / len1);
      double Exp2 = t.query(l2, r2, 1, n, 0).sum * (1.0 / len2);
      t.mult(l1, r1, 1.0 - (1.0 / (len1)), 1, n, 0);
      t.mult(l2, r2, 1.0 - (1.0 / (len2)), 1, n, 0);
      t.add(l1, r1, Exp2 * (1.0 / len1), 1, n, 0);
      t.add(l2, r2, Exp1 * (1.0 / len2), 1, n, 0);
    } else {
      int l, r;
      cin >> l >> r;
      cout << setprecision(9) << fixed << t.query(l, r, 1, n, 0).sum << '\n';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}