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 double eps = 1e-8;
double sg[100005 << 2];
double mtag[100005 << 2];
double ptag[100005 << 2];
double in[100005];
void build(int l, int r, int x) {
  if (l == r) return sg[x] = in[l], mtag[x] = 1, ptag[x] = 0, void();
  int m = l + r >> 1;
  build(l, m, (x + x + 1));
  build(m + 1, r, (x + x + 2));
  sg[x] = sg[(x + x + 1)] + sg[(x + x + 2)];
  mtag[x] = 1;
  ptag[x] = 0;
}
void push(int l, int r, int x, int i = 1) {
  if (l != r) {
    if (fabs(mtag[x] - 1) > eps) {
      sg[(x + x + 1)] *= mtag[x];
      sg[(x + x + 2)] *= mtag[x];
      mtag[(x + x + 1)] *= mtag[x];
      mtag[(x + x + 2)] *= mtag[x];
      ptag[(x + x + 1)] *= mtag[x];
      ptag[(x + x + 2)] *= mtag[x];
      mtag[x] = 1;
    }
    if (fabs(ptag[x]) > eps) {
      int m = l + r >> 1;
      sg[(x + x + 1)] += ptag[x] * (m - l + 1);
      sg[(x + x + 2)] += ptag[x] * (r - m);
      ptag[(x + x + 1)] += ptag[x];
      ptag[(x + x + 2)] += ptag[x];
      ptag[x] = 0;
    }
  }
}
void modify1(int l, int r, int ql, int qr, int x, double mul) {
  push(l, r, x);
  if (r < ql || l > qr) return;
  if (ql <= l && r <= qr) return sg[x] *= mul, mtag[x] *= mul, void();
  int m = l + r >> 1;
  modify1(l, m, ql, qr, (x + x + 1), mul);
  modify1(m + 1, r, ql, qr, (x + x + 2), mul);
  sg[x] = sg[(x + x + 1)] + sg[(x + x + 2)];
}
void modify2(int l, int r, int ql, int qr, int x, double pus) {
  push(l, r, x);
  if (r < ql || l > qr) return;
  if (ql <= l && r <= qr)
    return sg[x] += pus * (r - l + 1), ptag[x] += pus, void();
  int m = l + r >> 1;
  modify2(l, m, ql, qr, (x + x + 1), pus);
  modify2(m + 1, r, ql, qr, (x + x + 2), pus);
  sg[x] = sg[(x + x + 1)] + sg[(x + x + 2)];
}
double query(int l, int r, int ql, int qr, int x) {
  push(l, r, x);
  if (r < ql || l > qr) return 0;
  if (ql <= l && r <= qr) return sg[x];
  int m = l + r >> 1;
  return query(l, m, ql, qr, (x + x + 1)) +
         query(m + 1, r, ql, qr, (x + x + 2));
}
void brute(int l, int r, int x) {
  if (l != r) {
    push(l, r, x);
    int m = l + r >> 1;
    brute(l, m, (x + x + 1));
    brute(m + 1, r, (x + x + 2));
  }
}
int main() {
  {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
  };
  int n, q, cmd, l1, r1, l2, r2;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> in[i];
  build(1, n, 0);
  while (q--) {
    cin >> cmd;
    if (cmd == 1) {
      cin >> l1 >> r1 >> l2 >> r2;
      double pp = query(1, n, l2, r2, 0) / (r2 - l2 + 1) / (r1 - l1 + 1);
      double pp2 = query(1, n, l1, r1, 0) / (r2 - l2 + 1) / (r1 - l1 + 1);
      modify1(1, n, l1, r1, 0, (r1 - l1) * 1.0 / (r1 - l1 + 1));
      modify2(1, n, l1, r1, 0, pp);
      modify1(1, n, l2, r2, 0, (r2 - l2) * 1.0 / (r2 - l2 + 1));
      modify2(1, n, l2, r2, 0, pp2);
    } else {
      cin >> l1 >> r1;
      cout << fixed << setprecision(15) << query(1, n, l1, r1, 0) << '\n';
    }
  }
  return 0;
}