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;
int n, q;
int a[100007];
class Tree {
 public:
  double tr[5 * 100007];
  double lazymul[5 * 100007];
  double lazyadd[5 * 100007];
  void init(int where, int IL, int IR) {
    lazymul[where] = 1;
    lazyadd[where] = 0;
    if (IL == IR) {
      tr[where] = a[IL];
      return;
    }
    int mid = (IL + IR) / 2;
    init(2 * where, IL, mid);
    init(2 * where + 1, mid + 1, IR);
    merge(where);
  }
  void push_lazy(int where, int IL, int IR) {
    if (lazymul[where] != 1) {
      tr[where] = (tr[where] * lazymul[where]);
      if (IL != IR) {
        lazymul[2 * where] *= lazymul[where];
        lazyadd[2 * where] *= lazymul[where];
        lazymul[2 * where + 1] *= lazymul[where];
        lazyadd[2 * where + 1] *= lazymul[where];
      }
      lazymul[where] = 1;
    }
    if (lazyadd[where] != 0) {
      tr[where] = tr[where] + (IR - IL + 1) * lazyadd[where];
      if (IL != IR) {
        lazyadd[2 * where] += lazyadd[where];
        lazyadd[2 * where + 1] += lazyadd[where];
      }
      lazyadd[where] = 0;
    }
  }
  void merge(int where) {
    tr[where] = tr[2 * where] + tr[2 * where + 1];
    lazymul[where] = 1;
    lazyadd[where] = 0;
  }
  void update_mul(int where, int IL, int IR, int CURL, int CURR, double val) {
    push_lazy(where, IL, IR);
    if (IR < CURL || CURR < IL) {
      return;
    }
    if (CURL <= IL && IR <= CURR) {
      lazymul[where] *= val;
      push_lazy(where, IL, IR);
      return;
    }
    int mid = (IL + IR) / 2;
    update_mul(2 * where, IL, mid, CURL, CURR, val);
    update_mul(2 * where + 1, mid + 1, IR, CURL, CURR, val);
    merge(where);
  }
  void update_add(int where, int IL, int IR, int CURL, int CURR, double val) {
    push_lazy(where, IL, IR);
    if (IR < CURL || CURR < IL) {
      return;
    }
    if (CURL <= IL && IR <= CURR) {
      lazyadd[where] += val;
      push_lazy(where, IL, IR);
      return;
    }
    int mid = (IL + IR) / 2;
    update_add(2 * where, IL, mid, CURL, CURR, val);
    update_add(2 * where + 1, mid + 1, IR, CURL, CURR, val);
    merge(where);
  }
  double query(int where, int IL, int IR, int CURL, int CURR) {
    push_lazy(where, IL, IR);
    if (IR < CURL || CURR < IL) {
      return 0;
    }
    if (CURL <= IL && IR <= CURR) {
      return tr[where];
    }
    int mid = (IL + IR) / 2;
    return (query(2 * where, IL, mid, CURL, CURR) +
            query(2 * where + 1, mid + 1, IR, CURL, CURR));
  }
};
Tree w;
void input() {
  scanf("%d%d", &n, &q);
  int i;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  w.init(1, 1, n);
}
void solve() {
  int type;
  int x, y, z, t;
  while (q != 0) {
    q--;
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d%d%d", &x, &y, &z, &t);
      double sm1, sm2;
      sm1 = w.query(1, 1, n, x, y);
      sm2 = w.query(1, 1, n, z, t);
      double len1 = (y - x + 1);
      double len2 = (t - z + 1);
      w.update_mul(1, 1, n, x, y, (len1 - 1) / len1);
      w.update_mul(1, 1, n, z, t, (len2 - 1) / len2);
      w.update_add(1, 1, n, x, y, sm2 / (len2 * len1));
      w.update_add(1, 1, n, z, t, sm1 / (len1 * len2));
    } else {
      scanf("%d%d", &x, &y);
      printf("%.12f\n", w.query(1, 1, n, x, y));
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}