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 double a[4 * int(1e5 + 100)];
long double lazy[4 * int(1e5 + 100)];
long double lazy1[4 * int(1e5 + 100)];
void build(long long int b[], long long int l, long long int h,
           long long int pos) {
  if (l == h) {
    lazy1[pos - 1] = 1;
    a[pos - 1] = b[l - 1];
  } else {
    build(b, l, (l + h) / 2, 2 * pos);
    build(b, (l + h) / 2 + 1, h, 2 * pos + 1);
    a[pos - 1] = a[2 * pos - 1] + a[2 * pos];
    lazy1[pos - 1] = 1;
  }
}
long double query(long long int l, long long int h, long long int le,
                  long long int ri, long long int pos) {
  if (lazy[pos - 1]) {
    a[pos - 1] = lazy1[pos - 1] * a[pos - 1] + (h - l + 1) * lazy[pos - 1];
    if (l != h) {
      lazy[2 * pos - 1] = lazy[2 * pos - 1] * lazy1[pos - 1] + lazy[pos - 1];
      lazy[2 * pos] = lazy[2 * pos] * lazy1[pos - 1] + lazy[pos - 1];
      lazy1[2 * pos - 1] *= lazy1[pos - 1];
      lazy1[2 * pos] *= lazy1[pos - 1];
    }
    lazy[pos - 1] = 0;
    lazy1[pos - 1] = 1;
  }
  if (l >= le && h <= ri)
    return a[pos - 1];
  else if (ri < l || le > h)
    return 0;
  else {
    long double x = query(l, (l + h) / 2, le, ri, 2 * pos);
    long double y = query((l + h) / 2 + 1, h, le, ri, 2 * pos + 1);
    return (x + y);
  }
}
void updateutil(long long int l, long long int h, long long int le,
                long long int ri, long long int pos, long double x,
                long double m1) {
  if (lazy[pos - 1]) {
    a[pos - 1] = lazy1[pos - 1] * a[pos - 1] + (h - l + 1) * lazy[pos - 1];
    if (l != h) {
      lazy[2 * pos - 1] = lazy[2 * pos - 1] * lazy1[pos - 1] + lazy[pos - 1];
      lazy[2 * pos] = lazy[2 * pos] * lazy1[pos - 1] + lazy[pos - 1];
      lazy1[2 * pos - 1] *= lazy1[pos - 1];
      lazy1[2 * pos] *= lazy1[pos - 1];
    }
    lazy[pos - 1] = 0;
    lazy1[pos - 1] = 1;
  }
  if (ri < l || le > h)
    return;
  else if (l >= le && h <= ri) {
    a[pos - 1] = m1 * a[pos - 1] + (h - l + 1) * x;
    if (l != h) {
      lazy[2 * pos - 1] = lazy[2 * pos - 1] * m1 + x;
      lazy[2 * pos] = lazy[2 * pos] * m1 + x;
      lazy1[2 * pos - 1] *= m1;
      lazy1[2 * pos] *= m1;
    }
  } else {
    updateutil(l, (l + h) / 2, le, ri, 2 * pos, x, m1);
    updateutil((l + h) / 2 + 1, h, le, ri, 2 * pos + 1, x, m1);
    a[pos - 1] = a[2 * pos - 1] + a[2 * pos];
  }
}
void update(long long int n, long long int l1, long long int r1,
            long long int l2, long long int r2) {
  long long int n1 = (r1 - l1 + 1), n2 = (r2 - l2 + 1);
  long double x = query(1, n, l1, r1, 1), y = query(1, n, l2, r2, 1),
              m1 = (1.0 - 1.0 / n1), m2 = (1.0 - 1.0 / n2);
  updateutil(1, n, l1, r1, 1, y / (n1 * n2), m1);
  updateutil(1, n, l2, r2, 1, x / (n1 * n2), m2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, q, s, l1, r1, l2, r2;
  cin >> n >> q;
  long long int b[n];
  for (long long int i = 0; i < n; i++) cin >> b[i];
  build(b, 1, n, 1);
  while (q--) {
    cin >> s;
    if (s == 1) {
      cin >> l1 >> r1 >> l2 >> r2;
      update(n, l1, r1, l2, r2);
    } else {
      cin >> l1 >> r1;
      printf("%.6Lf\n", query(1, n, l1, r1, 1));
    }
  }
}