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 = 100010;
int n, q;
long double a[N];
long double d[N << 3];
long double b[N << 3];
long double val[N << 3];
void remake(int x) { val[x] = val[x + x] + val[x + x + 1]; }
void build(int x, int l, int r) {
  d[x] = 1;
  b[x] = 0;
  if (l == r) {
    val[x] = a[r];
    return;
  }
  build(x + x, l, ((l + r) >> 1));
  build(x + x + 1, ((l + r) >> 1) + 1, r);
  val[x] = val[x + x] + val[x + x + 1];
}
void push(int x, int l, int r) {
  if (d[x] != 1 || b[x] != 0) {
    val[x] = d[x] * val[x] + b[x] * 1LL * (r - l + 1);
    b[x + x] *= d[x];
    d[x + x] *= d[x];
    b[x + x + 1] *= d[x];
    d[x + x + 1] *= d[x];
    b[x + x] += b[x];
    b[x + x + 1] += b[x];
  }
  d[x] = 1;
  b[x] = 0;
}
void dv(int x, int l, int r, int u, int v, long double value) {
  push(x, l, r);
  if (l > v || r < u) return;
  if (l >= u && r <= v) {
    d[x] *= value;
    push(x, l, r);
    return;
  }
  dv(x + x, l, ((l + r) >> 1), u, v, value);
  dv(x + x + 1, ((l + r) >> 1) + 1, r, u, v, value);
  remake(x);
}
void add(int x, int l, int r, int u, int v, long double value) {
  push(x, l, r);
  if (l > v || r < u) return;
  if (l >= u && r <= v) {
    b[x] += value;
    return;
  }
  add(x + x, l, ((l + r) >> 1), u, v, value);
  add(x + x + 1, ((l + r) >> 1) + 1, r, u, v, value);
  remake(x);
}
long double query(int x, int l, int r, int u, int v) {
  push(x, l, r);
  if (l > v || r < u) return 0;
  if (l >= u && r <= v) return val[x];
  long double ans = query(x + x, l, ((l + r) >> 1), u, v) +
                    query(x + x + 1, ((l + r) >> 1) + 1, r, u, v);
  remake(x);
  return ans;
}
int main() {
  if (fopen("1.inp", "r")) {
    freopen("1.inp", "r", stdin);
  }
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  build(1, 1, n);
  cout << fixed << setprecision(7) << endl;
  while (q--) {
    int t, l, r;
    cin >> t >> l >> r;
    if (t == 2) {
      cout << query(1, 1, n, l, r) << '\n';
    } else {
      int u, v;
      cin >> u >> v;
      int m1 = r - l + 1, m2 = v - u + 1;
      long double one = query(1, 1, n, l, r);
      long double two = query(1, 1, n, u, v);
      dv(1, 1, n, l, r, 1LL * (m1 - 1) * m2);
      dv(1, 1, n, u, v, 1LL * (m2 - 1) * m1);
      add(1, 1, n, l, r, two);
      add(1, 1, n, u, v, one);
      dv(1, 1, n, l, r, 1.0 / (long double)(1LL * m1 * m2));
      dv(1, 1, n, u, v, 1.0 / (long double)(1LL * m1 * m2));
    }
  }
  return 0;
}