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 long long MOD = 1e9 + 7;
const long long MAXN = 2e5 + 7;
long double tree[MAXN * 4], a[MAXN], add[MAXN * 4], fact[MAXN * 4];
void build(long long v, long long tl, long long tr) {
  add[v] = 0;
  fact[v] = 1;
  if (tl == tr) {
    tree[v] = a[tl];
    return;
  }
  long long m = (tl + tr) / 2;
  build(v * 2 + 1, tl, m);
  build(v * 2 + 2, m + 1, tr);
  tree[v] = tree[v * 2 + 1] + tree[v * 2 + 2];
}
void push(long long v, long long tl, long long tr, long long m) {
  fact[v * 2 + 1] *= fact[v];
  add[v * 2 + 1] *= fact[v];
  tree[v * 2 + 1] *= fact[v];
  fact[v * 2 + 2] *= fact[v];
  add[v * 2 + 2] *= fact[v];
  tree[v * 2 + 2] *= fact[v];
  add[v * 2 + 1] += add[v];
  tree[v * 2 + 1] += add[v] * (m - tl + 1);
  add[v * 2 + 2] += add[v];
  tree[v * 2 + 2] += add[v] * (tr - m);
  add[v] = 0;
  fact[v] = 1;
}
void upd1(long long v, long long tl, long long tr, long long l, long long r,
          long double val) {
  if (tl > r || l > tr) return;
  if (tl >= l && tr <= r) {
    add[v] += val;
    tree[v] += val * (tr - tl + 1);
    return;
  }
  long long m = (tl + tr) / 2;
  push(v, tl, tr, m);
  upd1(v * 2 + 1, tl, m, l, r, val);
  upd1(v * 2 + 2, m + 1, tr, l, r, val);
  tree[v] = tree[v * 2 + 1] + tree[v * 2 + 2];
}
void upd2(long long v, long long tl, long long tr, long long l, long long r,
          long double val) {
  if (tl > r || l > tr) return;
  if (tl >= l && tr <= r) {
    add[v] *= val;
    fact[v] *= val;
    tree[v] *= val;
    return;
  }
  long long m = (tl + tr) / 2;
  push(v, tl, tr, m);
  upd2(v * 2 + 1, tl, m, l, r, val);
  upd2(v * 2 + 2, m + 1, tr, l, r, val);
  tree[v] = tree[v * 2 + 1] + tree[v * 2 + 2];
}
long double get(long long v, long long tl, long long tr, long long l,
                long long r) {
  if (tl > r || l > tr) return 0;
  if (tl >= l && tr <= r) return tree[v];
  long long m = (tl + tr) / 2;
  push(v, tl, tr, m);
  return get(v * 2 + 1, tl, m, l, r) + get(v * 2 + 2, m + 1, tr, l, r);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cout.precision(10);
  long long n, q;
  cin >> n >> q;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  build(0, 0, n - 1);
  for (long long i = 0; i < q; ++i) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      --l1;
      --l2;
      --r1;
      --r2;
      long double len1 = r1 - l1 + 1;
      long double len2 = r2 - l2 + 1;
      long double sr1 = get(0, 0, n - 1, l1, r1) / len1;
      long double sr2 = get(0, 0, n - 1, l2, r2) / len2;
      upd2(0, 0, n - 1, l1, r1, (len1 - 1) / len1);
      upd2(0, 0, n - 1, l2, r2, (len2 - 1) / len2);
      upd1(0, 0, n - 1, l1, r1, sr2 / len1);
      upd1(0, 0, n - 1, l2, r2, sr1 / len2);
    } else {
      long long l, r;
      cin >> l >> r;
      cout << get(0, 0, n - 1, l - 1, r - 1) << '\n';
    }
  }
  return 0;
}