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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
template <class T, class U>
inline void Max(T &a, U b) {
  if (a < b) a = b;
}
template <class T, class U>
inline void Min(T &a, U b) {
  if (a > b) a = b;
}
inline void add(int &a, int b) {
  a += b;
  while (a >= 1000000007) a -= 1000000007;
}
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = ans * (long long)a % 1000000007;
    a = (long long)a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
int a[100010];
double t[100010 << 2], w[100010 << 2], c[100010 << 2];
void up(int p) { t[p] = t[p << 1] + t[p << 1 | 1]; }
void build(int p, int l, int r) {
  w[p] = 1;
  if (l == r) {
    t[p] = a[l];
    return;
  }
  int m = (l + r) >> 1;
  build(p << 1, l, m);
  build(p << 1 | 1, m + 1, r);
  up(p);
}
void cal(int p, int l, int r, double v, double q) {
  t[p] = t[p] * q + v * (r - l + 1);
  w[p] *= q;
  c[p] *= q;
  c[p] += v;
}
void down(int p, int l, int r) {
  if (w[p] != 1 || c[p]) {
    int m = (l + r) >> 1;
    cal(p << 1, l, m, c[p], w[p]);
    cal(p << 1 | 1, m + 1, r, c[p], w[p]);
    c[p] = 0;
    w[p] = 1;
  }
}
void upd(int p, int l, int r, int first, int second, double v, double q) {
  if (l >= first && r <= second) {
    cal(p, l, r, v, q);
    return;
  }
  int m = (l + r) >> 1;
  down(p, l, r);
  if (first <= m) upd(p << 1, l, m, first, second, v, q);
  if (second > m) upd(p << 1 | 1, m + 1, r, first, second, v, q);
  up(p);
}
double query(int p, int l, int r, int first, int second) {
  if (l >= first && r <= second) {
    return t[p];
  }
  int m = (l + r) >> 1;
  down(p, l, r);
  double ans = 0;
  if (first <= m) ans = query(p << 1, l, m, first, second);
  if (second > m) ans += query(p << 1 | 1, m + 1, r, first, second);
  up(p);
  return ans;
}
int main() {
  cout << setprecision(9) << fixed;
  int j, i, k, T, ca = 0, K = 0, m = 0, n;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  build(1, 0, n - 1);
  int l1, r1, l2, r2;
  while (m--) {
    scanf("%d", &K);
    if (K == 1) {
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      l1--, r1--, l2--, r2--;
      double first = query(1, 0, n - 1, l1, r1),
             second = query(1, 0, n - 1, l2, r2);
      double q = 1. / (r1 - l1 + 1) / (r2 - l2 + 1);
      upd(1, 0, n - 1, l1, r1, second * q, 1 - 1. / (r1 - l1 + 1));
      upd(1, 0, n - 1, l2, r2, first * q, 1 - 1. / (r2 - l2 + 1));
    } else {
      scanf("%d%d", &l1, &r1);
      l1--, r1--;
      cout << query(1, 0, n - 1, l1, r1) << "\n";
    }
  }
  return 0;
}