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;
class customIO {
 public:
  istream& in;
  ostream& out;
  customIO(istream& _in, ostream& _out) : in(_in), out(_out) {}
  string next() {
    string s;
    in >> s;
    return s;
  }
  double nextDouble() { return stod(next()); }
  int nextInt() { return stoi(next()); }
  long long nextLong() { return stoll(next()); }
  template <typename T>
  void print(T t) {
    out << t;
  }
  template <typename T, typename... U>
  void print(T t, U... u) {
    out << t;
    print(u...);
  }
  template <typename T, typename... U>
  void println(T t, U... u) {
    print(t, u..., "\n");
  }
  void println() { print("\n"); }
};
struct node {
  bool f;
  double x, y;
};
int a[100005];
double tree[4 * 100002];
node lazy[4 * 100002];
void build(int node, int s, int e) {
  if (s > e) return;
  lazy[node].x = 1;
  lazy[node].y = 0;
  lazy[node].f = false;
  if (s == e) {
    tree[node] = a[s];
    return;
  }
  int mid = (s + e) / 2;
  build(2 * node, s, mid);
  build(2 * node + 1, mid + 1, e);
  tree[node] = tree[2 * node] + tree[2 * node + 1];
}
void update(int node, int s, int e, int l, int r, double p, double q) {
  if (s > e || l > r) return;
  if (lazy[node].f) {
    tree[node] = tree[node] * lazy[node].x + (e - s + 1) * lazy[node].y;
    if (s != e) {
      lazy[2 * node].x *= lazy[node].x;
      lazy[2 * node].y = lazy[2 * node].y * lazy[node].x + lazy[node].y;
      lazy[2 * node].f = true;
      lazy[2 * node + 1].x *= lazy[node].x;
      lazy[2 * node + 1].y = lazy[2 * node + 1].y * lazy[node].x + lazy[node].y;
      lazy[2 * node + 1].f = true;
    }
    lazy[node].f = false;
    lazy[node].x = 1;
    lazy[node].y = 0;
  }
  if (s > r || l > e) return;
  if (s >= l && e <= r) {
    tree[node] = tree[node] * p + (e - s + 1) * q;
    if (s != e) {
      lazy[2 * node].x *= p;
      lazy[2 * node].y = lazy[2 * node].y * p + q;
      lazy[2 * node].f = true;
      lazy[2 * node + 1].x *= p;
      lazy[2 * node + 1].y = lazy[2 * node + 1].y * p + q;
      lazy[2 * node + 1].f = true;
    }
    return;
  }
  int mid = (s + e) / 2;
  update(2 * node, s, mid, l, r, p, q);
  update(2 * node + 1, mid + 1, e, l, r, p, q);
  tree[node] = tree[2 * node] + tree[2 * node + 1];
}
double query(int node, int s, int e, int l, int r) {
  if (s > e || l > r) return 0;
  if (lazy[node].f) {
    tree[node] = tree[node] * lazy[node].x + (e - s + 1) * lazy[node].y;
    if (s != e) {
      lazy[2 * node].x *= lazy[node].x;
      lazy[2 * node].y = lazy[2 * node].y * lazy[node].x + lazy[node].y;
      lazy[2 * node].f = true;
      lazy[2 * node + 1].x *= lazy[node].x;
      lazy[2 * node + 1].y = lazy[2 * node + 1].y * lazy[node].x + lazy[node].y;
      lazy[2 * node + 1].f = true;
    }
    lazy[node].f = false;
    lazy[node].x = 1;
    lazy[node].y = 0;
  }
  if (s > r || l > e) return 0;
  if (s >= l && e <= r) {
    return tree[node];
  }
  int mid = (s + e) / 2;
  return query(2 * node, s, mid, l, r) + query(2 * node + 1, mid + 1, e, l, r);
}
class TaskE {
 public:
  void solve(customIO& io) {
    int n = io.nextInt();
    int q = io.nextInt();
    for (int i = 1; i <= n; ++i) {
      a[i] = io.nextInt();
    }
    build(1, 1, n);
    int ty, l1, r1, l2, r2;
    while (q--) {
      ty = io.nextInt();
      if (ty == 1) {
        l1 = io.nextInt();
        r1 = io.nextInt();
        l2 = io.nextInt();
        r2 = io.nextInt();
        double p = r1 - l1 + 1;
        double q = r2 - l2 + 1;
        double r = query(1, 1, n, l1, r1);
        double s = query(1, 1, n, l2, r2);
        update(1, 1, n, l1, r1, (p - 1) / p, s / (p * q));
        update(1, 1, n, l2, r2, (q - 1) / q, r / (p * q));
      } else {
        l1 = io.nextInt();
        r1 = io.nextInt();
        io.println(query(1, 1, n, l1, r1));
      }
    }
    return;
  }
};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout << std::fixed;
  std::cout.precision(20);
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  customIO io(in, out);
  TaskE solver;
  solver.solve(io);
  return 0;
}