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;
using db = double;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 5;
int n, m, opt, l[2], r[2];
int a[maxn];
int len[maxn << 4];
struct tree {
  db s, mul, add;
} T[maxn << 4];
void pup(int id) { T[id].s = T[id << 1].s + T[id << 1 | 1].s; }
void pdd(int id) {
  T[id << 1].s = T[id << 1].s * T[id].mul + T[id].add * len[id << 1];
  T[id << 1 | 1].s =
      T[id << 1 | 1].s * T[id].mul + T[id].add * len[id << 1 | 1];
  T[id << 1].add = T[id << 1].add * T[id].mul + T[id].add;
  T[id << 1 | 1].add = T[id << 1 | 1].add * T[id].mul + T[id].add;
  T[id << 1].mul = T[id << 1].mul * T[id].mul;
  T[id << 1 | 1].mul = T[id << 1 | 1].mul * T[id].mul;
  T[id].add = 0;
  T[id].mul = 1;
}
void build(int l, int r, int id) {
  T[id].mul = 1;
  T[id].add = 0;
  len[id] = r - l + 1;
  if (l == r) {
    T[id].s = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, id << 1);
  build(mid + 1, r, id << 1 | 1);
  pup(id);
}
void update(int L, int R, db d, db rat, int l, int r, int id) {
  if (L <= l && r <= R) {
    T[id].s = T[id].s * rat + d * len[id];
    T[id].add = T[id].add * rat + d;
    T[id].mul = T[id].mul * rat;
    return;
  }
  pdd(id);
  int mid = (l + r) / 2;
  if (mid >= L) update(L, R, d, rat, l, mid, id << 1);
  if (mid < R) update(L, R, d, rat, mid + 1, r, id << 1 | 1);
  pup(id);
}
db query(int L, int R, int l, int r, int id) {
  if (L <= l && r <= R) {
    return T[id].s;
  }
  pdd(id);
  int mid = (l + r) / 2;
  if (mid >= R)
    return query(L, R, l, mid, id << 1);
  else if (mid < L)
    return query(L, R, mid + 1, r, id << 1 | 1);
  else
    return query(L, R, l, mid, id << 1) + query(L, R, mid + 1, r, id << 1 | 1);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  build(1, n, 1);
  while (m--) {
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d%d%d%d", &l[0], &r[0], &l[1], &r[1]);
      db left = query(l[0], r[0], 1, n, 1);
      db right = query(l[1], r[1], 1, n, 1);
      update(l[0], r[0], right / (r[1] - l[1] + 1) / (r[0] - l[0] + 1),
             1.0 * (r[0] - l[0]) / (r[0] - l[0] + 1), 1, n, 1);
      update(l[1], r[1], left / (r[0] - l[0] + 1) / (r[1] - l[1] + 1),
             1.0 * (r[1] - l[1]) / (r[1] - l[1] + 1), 1, n, 1);
    } else {
      scanf("%d%d", &l[0], &r[0]);
      db res = query(l[0], r[0], 1, n, 1);
      printf("%.12lf\n", res);
    }
  }
  return 0;
}