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 long long N = 1e5 + 100;
long long a[N];
struct node {
  long double sum, toadd, tomul;
  node() {
    sum = 0;
    toadd = 0;
    tomul = 1;
  }
} seg[4 * N];
long long n, q;
void build(long long p = 1, long long s = 0, long long e = n) {
  if (e - s <= 1) {
    seg[p].sum = a[s];
    return;
  }
  long long mid = (s + e) / 2;
  build(2 * p, s, mid);
  build(2 * p + 1, mid, e);
  seg[p].sum = seg[2 * p].sum + seg[2 * p + 1].sum;
}
void mulit(long long p, long double val, long long s, long long e) {
  seg[p].sum *= val;
  seg[p].tomul *= val;
  seg[p].toadd *= val;
}
void addit(long long p, long double val, long long s, long long e) {
  seg[p].sum += (e - s) * val;
  seg[p].toadd += val;
}
void relax(long long p, long long s, long long e) {
  long double add = seg[p].toadd;
  long double mul = seg[p].tomul;
  seg[p].toadd = 0;
  seg[p].tomul = 1;
  long long mid = (s + e) / 2;
  mulit(2 * p, mul, s, mid);
  addit(2 * p, add, s, mid);
  mulit(2 * p + 1, mul, mid, e);
  addit(2 * p + 1, add, mid, e);
}
long double get(long long l, long long r, long long p = 1, long long s = 0,
                long long e = n) {
  if (s >= r || e <= l) return 0;
  if (s >= l && e <= r) {
    return seg[p].sum;
  }
  long long mid = (s + e) / 2;
  relax(p, s, e);
  return get(l, r, 2 * p, s, mid) + get(l, r, 2 * p + 1, mid, e);
}
void upd(long long l, long long r, long double vmul, long double vadd,
         long long p = 1, long long s = 0, long long e = n) {
  if (s >= r || e <= l) return;
  if (s >= l && e <= r) {
    mulit(p, vmul, s, e);
    addit(p, vadd, s, e);
    return;
  }
  long long mid = (s + e) / 2;
  relax(p, s, e);
  upd(l, r, vmul, vadd, 2 * p, s, mid);
  upd(l, r, vmul, vadd, 2 * p + 1, mid, e);
  seg[p].sum = seg[2 * p].sum + seg[2 * p + 1].sum;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  cout << fixed << setprecision(15);
  build();
  for (long long ii = 0, t; ii < q; ii++) {
    cin >> t;
    if (t == 1) {
      long long l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      l1--;
      l2--;
      long double s1 = get(l1, r1) / (long double)(r1 - l1);
      long double s2 = get(l2, r2) / (long double)(r2 - l2);
      long double val1 = (long double)(r1 - l1 - 1) / (long double)(r1 - l1);
      long double val2 = ((long double)1 / (long double)(r1 - l1)) * s2;
      upd(l1, r1, val1, val2);
      val1 = (long double)(r2 - l2 - 1) / (long double)(r2 - l2);
      val2 = ((long double)1 / (long double)(r2 - l2)) * s1;
      upd(l2, r2, val1, val2);
    } else {
      long long l, r;
      cin >> l >> r;
      l--;
      cout << get(l, r) << "\n";
    }
  }
  return 0;
}