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;
struct node {
  long double tagadd, tagmul;
  long double v;
  int len;
  void updatemul(long double x) {
    v *= x;
    tagmul *= x;
    tagadd *= x;
  }
  void updateadd(long double x) {
    tagadd += x;
    v += x * len;
  }
} f[N * 4];
int n, m;
int a[N];
void build(int l, int r, int s) {
  f[s].tagadd = 0;
  f[s].tagmul = 1;
  f[s].len = r - l + 1;
  if (l == r) {
    f[s].v = a[l];
    return;
  }
  build(l, (l + r) / 2, s + s);
  build((l + r) / 2 + 1, r, s + s + 1);
  f[s].v = f[s + s].v + f[s + s + 1].v;
}
void down(int l, int r, int s) {
  if (l != r) {
    f[s + s].updatemul(f[s].tagmul);
    f[s + s + 1].updatemul(f[s].tagmul);
    f[s + s].updateadd(f[s].tagadd);
    f[s + s + 1].updateadd(f[s].tagadd);
    f[s].tagmul = 1, f[s].tagadd = 0;
  }
}
void ins(int l, int r, int s, int ll, int rr, long double tagmul,
         long double tagadd) {
  down(l, r, s);
  if (r < ll || rr < l) return;
  if (ll <= l && r <= rr) {
    f[s].updatemul(tagmul);
    f[s].updateadd(tagadd);
    return;
  }
  ins(l, (l + r) / 2, s + s, ll, rr, tagmul, tagadd);
  ins((l + r) / 2 + 1, r, s + s + 1, ll, rr, tagmul, tagadd);
  f[s].v = f[s + s].v + f[s + s + 1].v;
}
long double get(int l, int r, int s, int ll, int rr) {
  down(l, r, s);
  if (r < ll || rr < l) return 0;
  if (ll <= l && r <= rr) {
    return f[s].v;
  }
  return get(l, (l + r) / 2, s + s, ll, rr) +
         get((l + r) / 2 + 1, r, s + s + 1, ll, rr);
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build(1, n, 1);
  while (m--) {
    int sig, l, r;
    scanf("%d %d %d", &sig, &l, &r);
    if (sig == 1) {
      int ll, rr;
      scanf("%d %d", &ll, &rr);
      long double all = get(1, n, 1, l, r), all1 = get(1, n, 1, ll, rr);
      int len = r - l + 1, len1 = rr - ll + 1;
      ins(1, n, 1, l, r, (long double)(len - 1) / len,
          (long double)all1 / (1ll * len1 * len));
      ins(1, n, 1, ll, rr, (long double)(len1 - 1) / len1,
          (long double)all / (1ll * len1 * len));
    } else {
      long double v = get(1, n, 1, l, r);
      printf("%.10f\n", (double)v);
    }
  }
  return 0;
}