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;
const int maxn = 1e5 + 10;
int a[maxn];
struct segtree {
  segtree *lc, *rc;
  double s, tagp, tags;
  void pull() { s = lc->s + rc->s; }
  segtree(int l, int r) {
    lc = rc = nullptr;
    s = 0.0;
    tagp = 1.0;
    tags = 0.0;
    if (l == r) {
      s = a[l] * 1.0;
      return;
    }
    lc = new segtree(l, ((l + r) >> 1));
    rc = new segtree(((l + r) >> 1) + 1, r);
    pull();
  }
  void modify(int l, int r, int ql, int qr, double v, double p) {
    push(l, r);
    if (l > qr || ql > r) return;
    if (l >= ql && r <= qr) {
      s = s * p + 1.0 * (r - l + 1) * v;
      tagp *= p;
      tags = tags * p + v;
      return;
    }
    lc->modify(l, ((l + r) >> 1), ql, qr, v, p);
    rc->modify(((l + r) >> 1) + 1, r, ql, qr, v, p);
    pull();
  }
  double query(int l, int r, int ql, int qr) {
    push(l, r);
    if (l > qr || ql > r) return 0.0;
    if (l >= ql && r <= qr) return s;
    return lc->query(l, ((l + r) >> 1), ql, qr) +
           rc->query(((l + r) >> 1) + 1, r, ql, qr);
  }
  void push(int l, int r) {
    if (l == r || tags == 0.0) return;
    lc->s = lc->s * tagp + 1.0 * (((l + r) >> 1) - l + 1) * tags;
    lc->tagp *= tagp;
    lc->tags = lc->tags * tagp + tags;
    rc->s = rc->s * tagp + 1.0 * (r - ((l + r) >> 1)) * tags;
    rc->tagp *= tagp;
    rc->tags = rc->tags * tagp + tags;
    tagp = 1.0;
    tags = 0.0;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  segtree *st = new segtree(1, n);
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      int len1 = r1 - l1 + 1, len2 = r2 - l2 + 1;
      double s1 = st->query(1, n, l1, r1);
      double s2 = st->query(1, n, l2, r2);
      s1 /= (len1 * 1.0 * len2);
      s2 /= (len1 * 1.0 * len2);
      double p1 = 1.0 * (r1 - l1) / (r1 - l1 + 1);
      double p2 = 1.0 * (r2 - l2) / (r2 - l2 + 1);
      st->modify(1, n, l1, r1, s2, p1);
      st->modify(1, n, l2, r2, s1, p2);
    } else {
      int l, r;
      cin >> l >> r;
      cout << fixed << setprecision(20) << st->query(1, n, l, r) << endl;
    }
  }
  return 0;
}