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 MXN = 2e5 + 10;
int n, q, A[MXN];
long double seg[MXN * 4];
pair<long double, long double> lazy[MXN * 4];
void Build(int id = 1, int l = 0, int r = n) {
  if (r - l == 1) {
    seg[id] = A[l];
    lazy[id] = {0, 1};
    return;
  }
  Build(id * 2, l, (l + r) / 2);
  Build(id * 2 + 1, (l + r) / 2, r);
  seg[id] = seg[id * 2] + seg[id * 2 + 1];
  lazy[id] = {0, 1};
}
void Shift(int id, int l, int r) {
  int mid = (l + r) / 2;
  long double a = lazy[id].first, b = lazy[id].second;
  lazy[id] = {0, 1};
  if (r - l == 1) return;
  seg[id * 2] = seg[id * 2] * b + a * (mid - l);
  seg[id * 2 + 1] = seg[id * 2 + 1] * b + a * (r - mid);
  lazy[id * 2].second *= b;
  lazy[id * 2 + 1].second *= b;
  lazy[id * 2].first = lazy[id * 2].first * b + a;
  lazy[id * 2 + 1].first = lazy[id * 2 + 1].first * b + a;
}
void Update(int le, int ri, long double val, long double rat, int id = 1,
            int l = 0, int r = n) {
  if (le >= r || ri <= l) return;
  if (le <= l && r <= ri) {
    Shift(id, l, r);
    seg[id] = seg[id] * rat + val * (r - l);
    lazy[id].second *= rat;
    lazy[id].first += val;
    return;
  }
  Shift(id, l, r);
  Update(le, ri, val, rat, id * 2, l, (l + r) / 2);
  Update(le, ri, val, rat, id * 2 + 1, (l + r) / 2, r);
  seg[id] = seg[id * 2] + seg[id * 2 + 1];
}
long double get(int le, int ri, int id = 1, int l = 0, int r = n) {
  if (le >= r || ri <= l) return (0);
  if (le <= l && r <= ri) return (seg[id]);
  Shift(id, l, r);
  return (get(le, ri, id * 2, l, (l + r) / 2) +
          get(le, ri, id * 2 + 1, (l + r) / 2, r));
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", &A[i]);
  Build();
  int tp, l, r, le, ri;
  long double cr, cl, rl, rr;
  while (q--) {
    scanf("%d%d%d", &tp, &l, &r);
    if (tp == 2) {
      printf("%.12lf\n", (double)get(l - 1, r));
      continue;
    }
    scanf("%d%d", &le, &ri);
    cl = get(l - 1, r) / (r - l + 1) / (ri - le + 1);
    cr = get(le - 1, ri) / (ri - le + 1) / (r - l + 1);
    rr = (ri - le);
    rr /= (ri - le + 1);
    rl = (r - l);
    rl /= (r - l + 1);
    Update(l - 1, r, cr, rl);
    Update(le - 1, ri, cl, rr);
  }
  return (0);
}