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;
long long n, g, t, l1, l2, r1, r2;
long long a[100005];
double seg[400005], lazyp[400005], lazyq[400005], p1, q1, p2, q2;
void build(long long i, long long l, long long r) {
  if (l == r) {
    seg[i] = (double)a[l];
    return;
  }
  long long mid = (l + r) / 2;
  build(2 * i, l, mid);
  build(2 * i + 1, mid + 1, r);
  seg[i] = seg[2 * i] + seg[2 * i + 1];
}
double query(long long i, long long l, long long r, long long x, long long y) {
  if ((lazyp[i] != (double)1) || lazyq[i]) {
    seg[i] = seg[i] * lazyp[i] + (double)(r - l + 1) * lazyq[i];
    if (l != r) {
      lazyp[2 * i] *= lazyp[i];
      lazyq[2 * i] = lazyq[2 * i] * lazyp[i] + lazyq[i];
      lazyp[2 * i + 1] *= lazyp[i];
      lazyq[2 * i + 1] = lazyq[2 * i + 1] * lazyp[i] + lazyq[i];
    }
    lazyp[i] = (double)1;
    lazyq[i] = (double)0;
  }
  if (l > y || x > r) return (double)0;
  if (l >= x && y >= r) return seg[i];
  long long mid = (l + r) / 2;
  return query(2 * i, l, mid, x, y) + query(2 * i + 1, mid + 1, r, x, y);
}
void update(long long i, long long l, long long r, long long x, long long y,
            double p, double q) {
  if ((lazyp[i] != (double)1) || lazyq[i]) {
    seg[i] = seg[i] * lazyp[i] + (double)(r - l + 1) * lazyq[i];
    if (l != r) {
      lazyp[2 * i] *= lazyp[i];
      lazyq[2 * i] = lazyq[2 * i] * lazyp[i] + lazyq[i];
      lazyp[2 * i + 1] *= lazyp[i];
      lazyq[2 * i + 1] = lazyq[2 * i + 1] * lazyp[i] + lazyq[i];
    }
    lazyp[i] = (double)1;
    lazyq[i] = (double)0;
  }
  if (l > y || x > r) return;
  if (l >= x && y >= r) {
    seg[i] = seg[i] * p + (double)(r - l + 1) * q;
    if (l != r) {
      lazyp[2 * i] *= p;
      lazyq[2 * i] = lazyq[2 * i] * p + q;
      lazyp[2 * i + 1] *= p;
      lazyq[2 * i + 1] = lazyq[2 * i + 1] * p + q;
    }
    return;
  }
  long long mid = (l + r) / 2;
  update(2 * i, l, mid, x, y, p, q);
  update(2 * i + 1, mid + 1, r, x, y, p, q);
  seg[i] = seg[2 * i] + seg[2 * i + 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (long long i = (long long)0; i < (long long)400005; i++)
    lazyp[i] = (double)1;
  cin >> n >> t;
  for (long long i = (long long)1; i < (long long)n + 1; i++) cin >> a[i];
  build(1, 1, n);
  while (t--) {
    cin >> g;
    if (g == 1) {
      cin >> l1 >> r1 >> l2 >> r2;
      p1 = (double)(r1 - l1) / (double)(r1 - l1 + 1);
      p2 = (double)(r2 - l2) / (double)(r2 - l2 + 1);
      q1 = query(1, 1, n, l2, r2) /
           ((double)(r1 - l1 + 1) * (double)(r2 - l2 + 1));
      q2 = query(1, 1, n, l1, r1) /
           ((double)(r1 - l1 + 1) * (double)(r2 - l2 + 1));
      update(1, 1, n, l1, r1, p1, q1);
      update(1, 1, n, l2, r2, p2, q2);
    } else {
      cin >> l1 >> r1;
      printf("%0.6lf\n", query(1, 1, n, l1, r1));
    }
  }
}