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;
const int N = 1e5 + 10;
int n, q, a[N];
namespace Seg {
int si[N << 2];
long double sm[N << 2], mt[N << 2], st[N << 2];
inline void pushup(int v) { sm[v] = sm[v << 1] + sm[v << 1 | 1]; }
inline void pushdown(int v) {
  if (mt[v] != 1) {
    sm[v] *= mt[v];
    if (si[v] > 1)
      mt[v << 1] *= mt[v], st[v << 1] *= mt[v], mt[v << 1 | 1] *= mt[v],
          st[v << 1 | 1] *= mt[v];
    mt[v] = 1;
  }
  if (st[v]) {
    sm[v] += si[v] * st[v];
    if (si[v] > 1) st[v << 1] += st[v], st[v << 1 | 1] += st[v];
    st[v] = 0;
  }
}
inline void init(int *a, int v = 1, int l = 1, int r = n) {
  si[v] = r - l + 1;
  mt[v] = 1;
  st[v] = 0;
  if (l == r) {
    sm[v] = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  init(a, v << 1, l, mid);
  init(a, v << 1 | 1, mid + 1, r);
  pushup(v);
}
inline void modi(int x, int y, long double z1, long double z2, int v = 1,
                 int l = 1, int r = n) {
  pushdown(v);
  if (l >= x && r <= y) {
    mt[v] *= z1;
    st[v] += z2;
    pushdown(v);
    return;
  }
  if (l > y || r < x) return;
  int mid = (l + r) >> 1;
  modi(x, y, z1, z2, v << 1, l, mid);
  modi(x, y, z1, z2, v << 1 | 1, mid + 1, r);
  pushup(v);
}
inline long double ask(int x, int y, int v = 1, int l = 1, int r = n) {
  if (l > y || r < x) return 0;
  pushdown(v);
  if (l >= x && r <= y) return sm[v];
  int mid = (l + r) >> 1;
  return ask(x, y, v << 1, l, mid) + ask(x, y, v << 1 | 1, mid + 1, r);
}
}  // namespace Seg
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  Seg::init(a);
  for (int i = 1, l, r, x, y, op; i <= q; ++i) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      scanf("%d%d", &l, &r);
      long double s1 = Seg::ask(x, y), s2 = Seg::ask(l, r);
      Seg::modi(x, y, 1 - double(1) / (y - x + 1),
                s2 / (r - l + 1) / (y - x + 1));
      Seg::modi(l, r, 1 - double(1) / (r - l + 1),
                s1 / (r - l + 1) / (y - x + 1));
    } else
      printf("%lf\n", (double)Seg::ask(x, y));
  }
  return 0;
}