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>
template <typename T>
inline void read(T &x) {
  char c;
  bool neg = false;
  while (!isdigit(c = getchar()) && c != '-')
    ;
  x = 0;
  if (c == '-') {
    neg = true;
    c = getchar();
  }
  do {
    x = x * 10 + c - '0';
  } while (isdigit(c = getchar()));
  if (neg) x = -x;
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    write(-x);
    return;
  }
  if (x < 10) {
    putchar(char(x + 48));
  } else {
    write(x / 10);
    putchar(char(48 + x % 10));
  }
}
template <typename T>
inline void writeln(T x) {
  write(x);
  putchar('\n');
}
using namespace std;
struct Data {
  double sum, product;
};
long long a[1001000];
void operator+=(Data &a, const Data &b) {
  if (b.product != 1) {
    a.product *= b.product;
    a.sum = a.sum * b.product;
  }
  if (b.sum != 0) a.sum += b.sum;
}
double IT[1001000];
Data lazy[1001000];
struct sIT {
  int n;
  void Build(int n) {
    this->n = n;
    Build(1, 1, n);
  }
  void Up(int l, int r, Data val) { Up(1, 1, n, l, r, val); }
  double Get(int l, int r) { return Get(1, 1, n, l, r); }

 private:
  void Build(int k, int l, int r) {
    if (l == r) {
      IT[k] = a[l];
      lazy[k].sum = 0;
      lazy[k].product = 1;
      return;
    }
    int m = (l + r) >> 1;
    Build(k * 2, l, m);
    Build(k * 2 + 1, m + 1, r);
    IT[k] = IT[k * 2] + IT[k * 2 + 1];
    lazy[k].sum = 0;
    lazy[k].product = 1;
  }
  void Down(int k, int l, int r) {
    int m = (l + r) >> 1;
    lazy[k * 2] += lazy[k];
    lazy[k * 2 + 1] += lazy[k];
    if (lazy[k].product != 1) {
      IT[k * 2] *= lazy[k].product;
      IT[k * 2 + 1] *= lazy[k].product;
    }
    if (lazy[k].sum != 0) {
      IT[k * 2] += lazy[k].sum * (m - l + 1);
      IT[k * 2 + 1] += lazy[k].sum * (r - m);
    }
    lazy[k].product = 1;
    lazy[k].sum = 0;
  }
  void Up(int k, int l, int r, int u, int v, Data val) {
    if (l > v || r < u) return;
    if (u <= l && r <= v) {
      if (val.product != 1) IT[k] *= val.product;
      if (val.sum != 0) IT[k] += val.sum * (r - l + 1);
      lazy[k] += val;
      return;
    }
    Down(k, l, r);
    int m = (l + r) >> 1;
    Up(k * 2, l, m, u, v, val);
    Up(k * 2 + 1, m + 1, r, u, v, val);
    IT[k] = IT[k * 2] + IT[k * 2 + 1];
  }
  double Get(int k, int l, int r, int u, int v) {
    if (l > v || r < u) return 0;
    if (u <= l && r <= v) return IT[k];
    Down(k, l, r);
    int m = (l + r) >> 1;
    return Get(k * 2, l, m, u, v) + Get(k * 2 + 1, m + 1, r, u, v);
  }
} ITT;
int n, q, c, l1, l2, r2, r1;
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  read(n);
  read(q);
  for (int i = 1; i <= n; i++) {
    read(a[i]);
  }
  ITT.Build(n);
  while (q--) {
    read(c);
    if (c == 1) {
      read(l1);
      read(r1);
      read(l2);
      read(r2);
      double sum1 = ITT.Get(l1, r1) / (r2 - l2 + 1),
             sum2 = ITT.Get(l2, r2) / (r2 - l2 + 1);
      ITT.Up(l1, r1,
             Data({sum2 / ((r1 - l1 + 1)), (double)(r1 - l1) / (r1 - l1 + 1)}));
      ITT.Up(l2, r2,
             Data({sum1 / ((r1 - l1 + 1)), (double)(r2 - l2) / (r2 - l2 + 1)}));
    } else {
      read(l1);
      read(r1);
      printf("%.6f\n", ITT.Get(l1, r1));
    }
  }
}