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;
double tree[400005], lazy[400005][2];
long long n, q, a[100005], i, t;
void build(int node, int start, int end) {
  if (start == end)
    tree[node] = a[start];
  else {
    int mid = (start + end) / 2;
    build(2 * node, start, mid);
    build(2 * node + 1, mid + 1, end);
    tree[node] = tree[2 * node] + tree[2 * node + 1];
  }
}
void update(int node, int start, int end, int l, int r, double m, double a) {
  if (lazy[node][0] != 1 || lazy[node][1] != 0) {
    tree[node] = tree[node] * lazy[node][0] + (end - start + 1) * lazy[node][1];
    if (start != end) {
      lazy[node * 2][0] *= lazy[node][0];
      lazy[node * 2 + 1][0] *= lazy[node][0];
      lazy[node * 2][1] = lazy[2 * node][1] * lazy[node][0] + lazy[node][1];
      lazy[node * 2 + 1][1] =
          lazy[2 * node + 1][1] * lazy[node][0] + lazy[node][1];
    }
    lazy[node][0] = 1;
    lazy[node][1] = 0;
  }
  if (start > end or start > r or end < l) return;
  if (start >= l and end <= r) {
    tree[node] = tree[node] * m + (end - start + 1) * a;
    if (start != end) {
      lazy[node * 2][0] *= m;
      lazy[node * 2 + 1][0] *= m;
      lazy[node * 2][1] = lazy[2 * node][1] * m + a;
      lazy[node * 2 + 1][1] = lazy[2 * node + 1][1] * m + a;
    }
    return;
  }
  int mid = (start + end) / 2;
  update(node * 2, start, mid, l, r, m, a);
  update(node * 2 + 1, mid + 1, end, l, r, m, a);
  tree[node] = tree[node * 2] + tree[node * 2 + 1];
}
double query(int node, int start, int end, int l, int r) {
  if (start > end or start > r or end < l) return 0;
  if (lazy[node][0] != 1 || lazy[node][1] != 0) {
    tree[node] = tree[node] * lazy[node][0] + (end - start + 1) * lazy[node][1];
    if (start != end) {
      lazy[node * 2][0] *= lazy[node][0];
      lazy[node * 2 + 1][0] *= lazy[node][0];
      lazy[node * 2][1] = lazy[2 * node][1] * lazy[node][0] + lazy[node][1];
      lazy[node * 2 + 1][1] =
          lazy[2 * node + 1][1] * lazy[node][0] + lazy[node][1];
    }
    lazy[node][0] = 1;
    lazy[node][1] = 0;
  }
  if (start >= l and end <= r) return tree[node];
  int mid = (start + end) / 2;
  double p1 = query(node * 2, start, mid, l, r);
  double p2 = query(node * 2 + 1, mid + 1, end, l, r);
  return (p1 + p2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(7);
  double l1, r1, l2, r2, m1, m2, a1, a2, l, r;
  for (i = 0; i <= 4e5; ++i) lazy[i][0] = 1;
  cin >> n >> q;
  for (i = 0; i < n; ++i) cin >> a[i];
  build(1, 0, n);
  while (q--) {
    cin >> t;
    if (t == 1) {
      cin >> l1 >> r1 >> l2 >> r2;
      m1 = (r1 - l1) / (r1 - l1 + 1);
      m2 = (r2 - l2) / (r2 - l2 + 1);
      a1 = query(1, 0, n, l2 - 1, r2 - 1) / ((r1 - l1 + 1) * (r2 - l2 + 1));
      a2 = query(1, 0, n, l1 - 1, r1 - 1) / ((r2 - l2 + 1) * (r1 - l1 + 1));
      update(1, 0, n, l1 - 1, r1 - 1, m1, a1);
      update(1, 0, n, l2 - 1, r2 - 1, m2, a2);
    } else {
      cin >> l >> r;
      cout << query(1, 0, n, l - 1, r - 1) << "\n";
    }
  }
  return 0;
}