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, a[100005];
double t[400005], mlz[400005], alz[400005];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = (x + (x << 2) << 1) + (ch ^ 48);
  return x * f;
}
void build(int i, int l, int r) {
  mlz[i] = 1;
  if (l == r) {
    t[i] = a[l];
    return;
  }
  int mid = l + r >> 1;
  build(i << 1, l, mid), build(i << 1 | 1, mid + 1, r),
      t[i] = t[i << 1] + t[i << 1 | 1];
}
void psd(int i, int l, int r) {
  int mid = l + r >> 1, li = i << 1, ri = li | 1;
  if (mlz[i] != 1)
    mlz[li] *= mlz[i], mlz[ri] *= mlz[i], alz[li] *= mlz[i], alz[ri] *= mlz[i],
        t[li] *= mlz[i], t[ri] *= mlz[i], mlz[i] = 1;
  if (alz[i])
    alz[li] += alz[i], alz[ri] += alz[i], t[li] += (mid - l + 1) * alz[i],
        t[ri] += (r - mid) * alz[i], alz[i] = 0;
}
double query(int i, int l, int r, int x, int y) {
  if (x <= l && y >= r) return t[i];
  psd(i, l, r);
  int mid = l + r >> 1;
  if (y <= mid) return query(i << 1, l, mid, x, y);
  if (x > mid) return query(i << 1 | 1, mid + 1, r, x, y);
  return query(i << 1, l, mid, x, y) + query(i << 1 | 1, mid + 1, r, x, y);
}
void mul(int i, int l, int r, int x, int y, double z) {
  if (x <= l && y >= r) {
    mlz[i] *= z, alz[i] *= z, t[i] *= z;
    return;
  }
  psd(i, l, r);
  int mid = l + r >> 1;
  if (x <= mid) mul(i << 1, l, mid, x, y, z);
  if (y > mid) mul(i << 1 | 1, mid + 1, r, x, y, z);
  t[i] = t[i << 1] + t[i << 1 | 1];
}
void add(int i, int l, int r, int x, int y, double z) {
  if (x <= l && y >= r) {
    alz[i] += z, t[i] += (r - l + 1) * z;
    return;
  }
  psd(i, l, r);
  int mid = l + r >> 1;
  if (x <= mid) add(i << 1, l, mid, x, y, z);
  if (y > mid) add(i << 1 | 1, mid + 1, r, x, y, z);
  t[i] = t[i << 1] + t[i << 1 | 1];
}
int main() {
  n = read(), q = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  build(1, 1, n);
  while (q--) {
    int fl = read(), x = read(), y = read();
    if (fl == 1) {
      int u = read(), v = read();
      double s = (double)(y - x + 1) * (v - u + 1),
             s1 = query(1, 1, n, x, y) / s, s2 = query(1, 1, n, u, v) / s;
      mul(1, 1, n, x, y, (double)(y - x) / (y - x + 1)),
          mul(1, 1, n, u, v, (double)(v - u) / (v - u + 1)),
          add(1, 1, n, x, y, s2), add(1, 1, n, u, v, s1);
    } else
      printf("%lf\n", query(1, 1, n, x, y));
  }
  return 0;
}