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 N = 1e5 + 10;
double seg[N << 2], lazyS[N << 2], lazyP[N << 2], val;
int n, q;
void Fine(int id, int lc, int s, int e) {
  seg[lc] *= lazyP[id], lazyP[lc] *= lazyP[id], lazyS[lc] *= lazyP[id];
  seg[lc] += (lazyS[id] * (double)(e - s)), lazyS[lc] += lazyS[id];
  return;
}
void relax(int id, int s, int mid, int e) {
  int lc = id * 2, rc = lc + 1;
  Fine(id, lc, s, mid);
  Fine(id, rc, mid, e);
  lazyP[id] = 1, lazyS[id] = 0;
}
void addS(int l, int r, int id = 1, int s = 0, int e = n) {
  if (r <= s || e <= l) return;
  if (l <= s && e <= r) {
    lazyS[id] += val, seg[id] += (val * (e - s));
    return;
  }
  int mid = (s + e) / 2, lc = id * 2, rc = lc + 1;
  relax(id, s, mid, e);
  addS(l, r, lc, s, mid);
  addS(l, r, rc, mid, e);
  seg[id] = seg[lc] + seg[rc];
  return;
}
void addP(int l, int r, int id = 1, int s = 0, int e = n) {
  if (r <= s || e <= l) return;
  if (l <= s && e <= r) {
    seg[id] *= val, lazyP[id] *= val, lazyS[id] *= val;
    return;
  }
  int mid = (s + e) / 2, lc = id * 2, rc = lc + 1;
  relax(id, s, mid, e);
  addP(l, r, lc, s, mid);
  addP(l, r, rc, mid, e);
  seg[id] = seg[lc] + seg[rc];
  return;
}
double get(int l, int r, int id = 1, int s = 0, int e = n) {
  if (r <= s || e <= l) return 0;
  if (l <= s && e <= r) return seg[id];
  int mid = (s + e) / 2, lc = id * 2, rc = lc + 1;
  relax(id, s, mid, e);
  return get(l, r, lc, s, mid) + get(l, r, rc, mid, e);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(7);
  cin >> n >> q;
  memset(lazyP, 1, sizeof(lazyP));
  for (int i = 0; i < n; i++) {
    cin >> val;
    addS(i, i + 1);
  }
  for (int i = 0; i < q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int l1, l2, r1, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      l1--, l2--;
      double rn1 = get(l1, r1), rn2 = get(l2, r2);
      rn1 = (double)rn1 / (double)(r1 - l1) / (double)(r2 - l2);
      rn2 = (double)rn2 / (double)(r2 - l2) / (double)(r1 - l1);
      val = (double)(r2 - l2 - 1) / (double)(r2 - l2);
      addP(l2, r2);
      val = (double)(r1 - l1 - 1) / (double)(r1 - l1);
      addP(l1, r1);
      val = rn1;
      addS(l2, r2);
      val = rn2;
      addS(l1, r1);
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      cout << get(l, r) << "\n";
      ;
    }
  }
  return 0;
}