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 + 4, lim = 5 * maxn;
const long double eps = 1e-14;
long double segtree[lim], nochgprob[lim], change[lim], ip[maxn];
int n, q;
void build(int cur, int l, int r) {
  nochgprob[cur] = 1;
  change[cur] = 0;
  if (l == r) {
    segtree[cur] = ip[l];
    return;
  }
  int lc = 2 * cur, rc = lc + 1, mid = (l + r) / 2;
  build(lc, l, mid);
  build(rc, mid + 1, r);
  segtree[cur] = segtree[lc] + segtree[rc];
}
void pushdown(int cur, int lc, int rc, int l, int mid, int r) {
  segtree[lc] = segtree[lc] * nochgprob[cur] + (mid - l + 1) * change[cur];
  nochgprob[lc] *= nochgprob[cur];
  change[lc] *= nochgprob[cur];
  change[lc] += change[cur];
  segtree[rc] = segtree[rc] * nochgprob[cur] + (r - mid) * change[cur];
  nochgprob[rc] *= nochgprob[cur];
  change[rc] *= nochgprob[cur];
  change[rc] += change[cur];
  nochgprob[cur] = 1;
  change[cur] = 0;
}
long double sum(int cur, int l, int r, int ql, int qr) {
  if (r < ql || l > qr) return 0;
  if (l >= ql && r <= qr) return segtree[cur];
  int lc = 2 * cur, rc = lc + 1, mid = (l + r) / 2;
  if (abs(nochgprob[cur] - 1) > eps || abs(change[cur]) > eps)
    pushdown(cur, lc, rc, l, mid, r);
  return (sum(lc, l, mid, ql, qr) + sum(rc, mid + 1, r, ql, qr));
}
long double ncp, cp;
void update(int cur, int l, int r, int ql, int qr) {
  if (l > qr || r < ql) return;
  if (l >= ql && r <= qr) {
    segtree[cur] = segtree[cur] * ncp + (r - l + 1) * cp;
    nochgprob[cur] *= ncp;
    change[cur] *= ncp;
    change[cur] += cp;
    return;
  }
  int lc = 2 * cur, rc = lc + 1, mid = (l + r) / 2;
  if (abs(nochgprob[cur] - 1) > eps || abs(change[cur]) > eps)
    pushdown(cur, lc, rc, l, mid, r);
  update(lc, l, mid, ql, qr);
  update(rc, mid + 1, r, ql, qr);
  segtree[cur] = segtree[lc] + segtree[rc];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> ip[i];
  build(1, 1, n);
  cout.precision(7);
  while (q--) {
    int t, l1, r1, l2, r2;
    cin >> t;
    if (t == 2) {
      cin >> l1 >> r1;
      cout << fixed << sum(1, 1, n, l1, r1) << '\n';
    } else {
      cin >> l1 >> r1 >> l2 >> r2;
      long double sum1 = sum(1, 1, n, l1, r1), sum2 = sum(1, 1, n, l2, r2);
      ncp = (r1 - l1);
      ncp /= (r1 - l1 + 1);
      cp = sum2 / (r2 - l2 + 1);
      cp /= (r1 - l1 + 1);
      update(1, 1, n, l1, r1);
      ncp = (r2 - l2);
      ncp /= (r2 - l2 + 1);
      cp = sum1 / (r1 - l1 + 1);
      cp /= (r2 - l2 + 1);
      update(1, 1, n, l2, r2);
    }
  }
  return 0;
}