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 oo = 0x7fffffff;
const long long OO = 0x7fffffffffffffff;
const int NX = 1e5 + 100;
long double tree[4 * NX], mtree[4 * NX], stree[4 * NX], arr[NX];
int le[4 * NX], ri[4 * NX], N, Q;
void init(int i) {
  mtree[i] = 1;
  stree[i] = 0;
  int m = (le[i] + ri[i]) / 2;
  if (le[i] < ri[i]) {
    le[i * 2] = le[i];
    ri[i * 2] = m;
    le[i * 2 + 1] = m + 1;
    ri[i * 2 + 1] = ri[i];
    init(i * 2);
    init(i * 2 + 1);
    tree[i] = tree[i * 2] + tree[i * 2 + 1];
  } else
    tree[i] = arr[m];
}
void propagate(int i) {
  tree[i] = tree[i] * mtree[i] + stree[i] * (ri[i] - le[i] + 1);
  if (le[i] < ri[i]) {
    mtree[i * 2] *= mtree[i];
    mtree[i * 2 + 1] *= mtree[i];
    stree[i * 2] = stree[i * 2] * mtree[i] + stree[i];
    stree[i * 2 + 1] = stree[i * 2 + 1] * mtree[i] + stree[i];
  }
  mtree[i] = 1;
  stree[i] = 0;
}
long double rmq_query(int l, int r, int i) {
  propagate(i);
  if (r < le[i] || ri[i] < l) return 0;
  if (l <= le[i] && ri[i] <= r) return tree[i];
  return rmq_query(l, r, i * 2) + rmq_query(l, r, i * 2 + 1);
}
void rmq_update(int l, int r, long double M, long double S, int i) {
  propagate(i);
  if (r < le[i] || ri[i] < l) return;
  if (l <= le[i] && ri[i] <= r) {
    stree[i] = S;
    mtree[i] = M;
    propagate(i);
    return;
  }
  rmq_update(l, r, M, S, i * 2);
  rmq_update(l, r, M, S, i * 2 + 1);
  tree[i] = tree[2 * i] + tree[i * 2 + 1];
}
long double query(int l, int r) { return rmq_query(l, r, 1); }
void update(int l1, int r1, int l2, int r2) {
  long double s1 = query(l1, r1), s2 = query(l2, r2), f1 = r1 - l1 + 1,
              f2 = r2 - l2 + 1;
  rmq_update(l1, r1, (f1 - 1) / f1, s2 / (f1 * f2), 1);
  rmq_update(l2, r2, (f2 - 1) / f2, s1 / (f1 * f2), 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N >> Q;
  for (int i = 1; i <= N; i++) cin >> arr[i];
  le[1] = 1;
  ri[1] = N;
  init(1);
  while (Q--) {
    int com, l1, l2, r1, r2;
    cin >> com >> l1 >> r1;
    if (com == 1) {
      cin >> l2 >> r2;
      update(l1, r1, l2, r2);
    } else {
      cout << query(l1, r1) << "\n";
    }
  }
  return 0;
}