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;
int const N = 1e5 + 5;
int n, q;
int a[N];
double tree[8 * N];
pair<double, double> add[8 * N];
void build(int node, int l, int r) {
  add[node] = make_pair(1, 0);
  if (l == r) {
    tree[node] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build(node * 2, l, mid);
  build(node * 2 + 1, mid + 1, r);
  tree[node] = tree[2 * node] + tree[2 * node + 1];
}
void Plus(pair<double, double> &p, pair<double, double> q) {
  p.first *= q.first;
  p.second = p.second * q.first + q.second;
}
void down(int node, int l, int r) {
  int mid = (l + r) / 2;
  tree[2 * node] =
      tree[2 * node] * add[node].first + add[node].second * (mid - l + 1);
  tree[2 * node + 1] =
      tree[2 * node + 1] * add[node].first + add[node].second * (r - mid);
  Plus(add[2 * node], add[node]);
  Plus(add[2 * node + 1], add[node]);
  add[node] = make_pair(1, 0);
}
void update(int node, int l, int r, int d, int c, pair<double, double> val) {
  if (l > c || d > r) {
    return;
  }
  if (d <= l && r <= c) {
    tree[node] = tree[node] * val.first + val.second * (r - l + 1);
    Plus(add[node], val);
    return;
  }
  down(node, l, r);
  int mid = (l + r) / 2;
  update(node * 2, l, mid, d, c, val);
  update(node * 2 + 1, mid + 1, r, d, c, val);
  tree[node] = tree[2 * node] + tree[2 * node + 1];
}
double query(int node, int l, int r, int d, int c) {
  if (l > c || r < d) return 0;
  if (d <= l && r <= c) return tree[node];
  down(node, l, r);
  int mid = (l + r) / 2;
  return query(2 * node, l, mid, d, c) + query(2 * node + 1, mid + 1, r, d, c);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  build(1, 1, n);
  int l1, r1, l2, r2, type;
  while (q--) {
    cin >> type;
    if (type == 1) {
      cin >> l1 >> r1 >> l2 >> r2;
      double sum1 = query(1, 1, n, l1, r1);
      double sum2 = query(1, 1, n, l2, r2);
      double size1 = r1 - l1 + 1;
      double size2 = r2 - l2 + 1;
      update(1, 1, n, l1, r1,
             make_pair((size1 - 1) / size1, sum2 / (size1 * size2)));
      update(1, 1, n, l2, r2,
             make_pair((size2 - 1) / size2, sum1 / (size1 * size2)));
    } else {
      cin >> l1 >> r1;
      printf("%0.6f\n", query(1, 1, n, l1, r1));
    }
  }
}