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;
struct {
  long double sum, lazy_prod, lazy_sum;
} tree[100005 * 4];
int a[100005];
void update_node(int idx, int b, int e, long double prod, long double sum) {
  tree[idx].sum = prod * tree[idx].sum + sum * (long double)(e - b + 1);
  tree[idx].lazy_prod *= prod;
  tree[idx].lazy_sum = tree[idx].lazy_sum * prod + sum;
}
void update_lazy(int idx, int b, int e) {
  update_node((idx * 2), (b), ((b + e) / 2), tree[idx].lazy_prod,
              tree[idx].lazy_sum);
  update_node((idx * 2 + 1), ((b + e) / 2 + 1), (e), tree[idx].lazy_prod,
              tree[idx].lazy_sum);
  tree[idx].lazy_prod = 1.0;
  tree[idx].lazy_sum = 0.0;
}
void init(int idx, int b, int e) {
  if (b == e) {
    tree[idx].lazy_prod = 1.0;
    tree[idx].sum = a[b];
    return;
  }
  init((idx * 2), (b), ((b + e) / 2));
  init((idx * 2 + 1), ((b + e) / 2 + 1), (e));
  tree[idx].sum = tree[idx * 2].sum + tree[(idx * 2) + 1].sum;
  tree[idx].lazy_prod = 1.0;
}
void update(int idx, int b, int e, int &st, int &en, long double &prod,
            long double &sum) {
  if (st > e || en < b) return;
  if (st <= b && e <= en) {
    update_node(idx, b, e, prod, sum);
    return;
  }
  update_lazy(idx, b, e);
  update((idx * 2), (b), ((b + e) / 2), st, en, prod, sum);
  update((idx * 2 + 1), ((b + e) / 2 + 1), (e), st, en, prod, sum);
  tree[idx].sum = tree[idx * 2].sum + tree[(idx * 2) + 1].sum;
}
long double query(int idx, int b, int e, int &st, int &en) {
  if (st > e || en < b) return 0.0;
  if (st <= b && e <= en) return tree[idx].sum;
  update_lazy(idx, b, e);
  return query((idx * 2), (b), ((b + e) / 2), st, en) +
         query((idx * 2 + 1), ((b + e) / 2 + 1), (e), st, en);
}
int main() {
  int n, i, q, k, l1, l2, r1, r2;
  long double sum1, sum2, prod, sum;
  scanf("%d %d", &n, &q);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  init(1, 1, n);
  while (q--) {
    scanf("%d", &k);
    if (k == 1) {
      scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
      sum1 = query(1, 1, n, l1, r1);
      sum2 = query(1, 1, n, l2, r2);
      prod = ((long double)(r1 - l1)) / ((long double)(r1 - l1 + 1));
      sum = (1.0 / ((long double)(r1 - l1 + 1))) *
            (1.0 / ((long double)(r2 - l2 + 1))) * sum2;
      update(1, 1, n, l1, r1, prod, sum);
      prod = ((long double)(r2 - l2)) / ((long double)(r2 - l2 + 1));
      sum = (1.0 / ((long double)(r1 - l1 + 1))) *
            (1.0 / ((long double)(r2 - l2 + 1))) * sum1;
      update(1, 1, n, l2, r2, prod, sum);
    } else {
      scanf("%d %d", &l1, &r1);
      printf("%0.6Lf\n", query(1, 1, n, l1, r1));
    }
  }
  return 0;
}