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;
using LL = long long;
void Print() { cout << endl; }
template <typename T1, typename... T>
void Print(const T1 &t1, const T &...t) {
  cout << t1 << " ";
  Print(t...);
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <class Iterator>
ostream &output_cantainer(ostream &os, Iterator first, Iterator last,
                          bool space = true) {
  if (space) {
    for (Iterator it = first; it != last; it++) {
      os << (*it) << ' ';
    }
    os << endl;
  } else
    for (Iterator it = first; it != last; it++) {
      os << (*it) << endl;
    }
  return os;
}
const int N = 1e5 + 4;
int a[N], n, q;
double k[N << 2], b[N << 2], sum[N << 2], siz[N << 2];
void push_down(int id) {
  if (k[id] != 1.0 || b[id] != 0.0) {
    double lk = k[id << 1], lb = b[id << 1];
    double rk = k[id << 1 | 1], rb = b[id << 1 | 1];
    k[id << 1] = lk * k[id];
    k[id << 1 | 1] = rk * k[id];
    b[id << 1] = k[id] * lb + b[id];
    b[id << 1 | 1] = k[id] * rb + b[id];
    sum[id << 1] = k[id] * sum[id << 1] + siz[id << 1] * b[id];
    sum[id << 1 | 1] = k[id] * sum[id << 1 | 1] + siz[id << 1 | 1] * b[id];
    k[id] = 1.0;
    b[id] = 0.0;
  }
}
void push_up(int id) { sum[id] = sum[id << 1] + sum[id << 1 | 1]; }
void build(int id, int l, int r) {
  k[id] = 1.0;
  b[id] = 0.0;
  if (l == r) {
    sum[id] = a[l];
    siz[id] = 1;
    return;
  }
  int m = (l + r) >> 1;
  build(id << 1, l, m);
  build(id << 1 | 1, m + 1, r);
  siz[id] = r - l + 1;
  push_up(id);
}
void update(int id, int l, int r, int x, int y, double k1, double b1) {
  if (r < x || l > y) return;
  if (x <= l && r <= y) {
    double ok = k[id], ob = b[id];
    sum[id] = sum[id] * k1 + b1 * siz[id];
    k[id] = k1 * ok;
    b[id] = k1 * ob + b1;
    return;
  }
  push_down(id);
  int m = (l + r) >> 1;
  update(id << 1, l, m, x, y, k1, b1);
  update(id << 1 | 1, m + 1, r, x, y, k1, b1);
  push_up(id);
}
double query(int id, int l, int r, int x, int y) {
  if (r < x || l > y) return 0.0;
  if (x <= l && r <= y) return sum[id];
  push_down(id);
  int m = (l + r) >> 1;
  return query(id << 1, l, m, x, y) + query(id << 1 | 1, m + 1, r, x, y);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  build(1, 1, n);
  while (q--) {
    int t, l1, r1, l2, r2;
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      int len1 = r1 - l1 + 1, len2 = r2 - l2 + 1;
      double s1 = query(1, 1, n, l1, r1) / len1;
      double s2 = query(1, 1, n, l2, r2) / len2;
      update(1, 1, n, l1, r1, 1.0 * (len1 - 1) / len1, 1.0 / len1 * s2);
      update(1, 1, n, l2, r2, 1.0 * (len2 - 1) / len2, 1.0 / len2 * s1);
    } else {
      scanf("%d%d", &l1, &r1);
      printf("%.8f\n", query(1, 1, n, l1, r1));
    }
  }
}