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>
int inp() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return sum;
}
double a[100010];
struct SEG_Tree {
  struct SEG {
    int l;
    int r;
    double lazy2;
    double sum;
    double lazy;
  } t[800010];
  void push_down(int cur) {
    if (t[cur].lazy2 != 1) {
      t[((cur) << 1)].lazy2 *= t[cur].lazy2;
      t[((cur) << 1 | 1)].lazy2 *= t[cur].lazy2;
      t[((cur) << 1)].lazy *= t[cur].lazy2;
      t[((cur) << 1 | 1)].lazy *= t[cur].lazy2;
      t[((cur) << 1)].sum *= t[cur].lazy2;
      t[((cur) << 1 | 1)].sum *= t[cur].lazy2;
      t[cur].lazy2 = 1;
    }
    t[((cur) << 1)].sum +=
        t[cur].lazy * (double)(t[((cur) << 1)].r - t[((cur) << 1)].l + 1);
    t[((cur) << 1 | 1)].sum +=
        t[cur].lazy *
        (double)(t[((cur) << 1 | 1)].r - t[((cur) << 1 | 1)].l + 1);
    t[((cur) << 1)].lazy += t[cur].lazy;
    t[((cur) << 1 | 1)].lazy += t[cur].lazy;
    t[cur].lazy = 0;
  }
  void push_up(int cur) {
    t[cur].sum = t[((cur) << 1)].sum + t[((cur) << 1 | 1)].sum;
  }
  void build(int cur, int l, int r) {
    t[cur].l = l;
    t[cur].r = r;
    t[cur].lazy = 0;
    t[cur].lazy2 = 1;
    if (l == r) {
      t[cur].sum = a[l];
      return;
    }
    int mid = (l + r) >> 1;
    build(((cur) << 1), l, mid);
    build(((cur) << 1 | 1), mid + 1, r);
    push_up(cur);
  }
  void clear(int cur, int l, int r, double c) {
    if (t[cur].l > r || t[cur].r < l) return;
    push_down(cur);
    if (t[cur].l >= l && t[cur].r <= r) {
      t[cur].lazy2 *= c;
      t[cur].lazy *= c;
      t[cur].sum *= c;
      return;
    }
    clear(((cur) << 1), l, r, c);
    clear(((cur) << 1 | 1), l, r, c);
    push_up(cur);
  }
  void modify(int cur, int l, int r, double c) {
    if (t[cur].l > r || t[cur].r < l) return;
    push_down(cur);
    if (t[cur].l >= l && t[cur].r <= r) {
      t[cur].lazy += c;
      t[cur].sum += (double)(t[cur].r - t[cur].l + 1) * c;
      return;
    }
    modify(((cur) << 1), l, r, c);
    modify(((cur) << 1 | 1), l, r, c);
    push_up(cur);
  }
  double query(int cur, int l, int r) {
    if (t[cur].l > r || t[cur].r < l) return 0;
    push_down(cur);
    if (t[cur].l >= l && t[cur].r <= r) return t[cur].sum;
    return query(((cur) << 1), l, r) + query(((cur) << 1 | 1), l, r);
  }
} t;
int main() {
  int n = inp();
  int m = inp();
  for (int i = 1; i <= n; i++) a[i] = inp();
  t.build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int op = inp();
    if (op == 1) {
      int l1 = inp();
      int r1 = inp();
      int l2 = inp();
      int r2 = inp();
      double sum1 = t.query(1, l1, r1);
      double sum2 = t.query(1, l2, r2);
      sum1 /= (double)(r1 - l1 + 1);
      sum2 /= (double)(r2 - l2 + 1);
      t.clear(1, l1, r1, (double)(r1 - l1) / (double)(r1 - l1 + 1));
      t.modify(1, l1, r1, sum2 / (double)(r1 - l1 + 1));
      t.clear(1, l2, r2, (double)(r2 - l2) / (double)(r2 - l2 + 1));
      t.modify(1, l2, r2, sum1 / (double)(r2 - l2 + 1));
    } else {
      int l = inp();
      int r = inp();
      printf("%.7lf\n", t.query(1, l, r));
    }
  }
}