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-9;
const int M = 1 << 17;
int n, q;
int a[M];
double mul[2 * M];
double add[2 * M];
double sub[2 * M];
void push(int v, int l, int r) {
  if (mul[v] > 1 + eps || mul[v] < 1 - eps) {
    sub[2 * v] *= mul[v];
    sub[2 * v + 1] *= mul[v];
    mul[2 * v] *= mul[v];
    mul[2 * v + 1] *= mul[v];
    add[2 * v] *= mul[v];
    add[2 * v + 1] *= mul[v];
    mul[v] = 1;
  }
  if (add[v] > eps || add[v] < -eps) {
    sub[2 * v] += add[v] * (r - l + 1) / 2;
    sub[2 * v + 1] += add[v] * (r - l + 1) / 2;
    add[2 * v] += add[v];
    add[2 * v + 1] += add[v];
    add[v] = 0;
  }
}
void build(int v = 1, int l = 0, int r = M - 1) {
  mul[v] = 1;
  if (l == r) {
    sub[v] = a[l];
    return;
  }
  build(2 * v, l, (l + r) / 2);
  build(2 * v + 1, (l + r) / 2 + 1, r);
  sub[v] = sub[2 * v] + sub[2 * v + 1];
}
double query(int a, int b, int v = 1, int l = 0, int r = M - 1) {
  if (a > b || l > b || a > r) return 0;
  if (a <= l && r <= b) return sub[v];
  push(v, l, r);
  double ans = query(a, b, 2 * v, l, (l + r) / 2) +
               query(a, b, 2 * v + 1, (l + r) / 2 + 1, r);
  sub[v] = sub[2 * v] + sub[2 * v + 1];
  return ans;
}
void insert_mul(int a, int b, double c, int v = 1, int l = 0, int r = M - 1) {
  if (a > b || l > b || a > r) return;
  if (a <= l && r <= b) {
    sub[v] *= c;
    mul[v] *= c;
    add[v] *= c;
    return;
  }
  push(v, l, r);
  insert_mul(a, b, c, 2 * v, l, (l + r) / 2);
  insert_mul(a, b, c, 2 * v + 1, (l + r) / 2 + 1, r);
  sub[v] = sub[2 * v] + sub[2 * v + 1];
}
void insert_add(int a, int b, double c, int v = 1, int l = 0, int r = M - 1) {
  if (a > b || l > b || a > r) return;
  if (a <= l && r <= b) {
    sub[v] += (r - l + 1) * c;
    add[v] += c;
    return;
  }
  push(v, l, r);
  insert_add(a, b, c, 2 * v, l, (l + r) / 2);
  insert_add(a, b, c, 2 * v + 1, (l + r) / 2 + 1, r);
  sub[v] = sub[2 * v] + sub[2 * v + 1];
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  build();
  cout << fixed << setprecision(10);
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int a, b, c, d;
      cin >> a >> b >> c >> d;
      double avgab = query(a, b) / (b - a + 1);
      double avgcd = query(c, d) / (d - c + 1);
      insert_mul(a, b, (double)(b - a) / (b - a + 1));
      insert_mul(c, d, (double)(d - c) / (d - c + 1));
      insert_add(a, b, (double)1 / (b - a + 1) * avgcd);
      insert_add(c, d, (double)1 / (d - c + 1) * avgab);
    } else {
      int a, b;
      cin >> a >> b;
      cout << query(a, b) << '\n';
    }
  }
}