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 N = 1e5 + 5;
double ary[N], tree[4 * N], lazy_mul[4 * N], lazy_add[4 * N];
int n, q;
void push(int node, int s, int e) {
  if (s != e) {
    lazy_mul[(node << 1)] *= lazy_mul[node];
    lazy_mul[(node << 1) + 1] *= lazy_mul[node];
    lazy_add[(node << 1)] =
        lazy_add[(node << 1)] * lazy_mul[node] + lazy_add[node];
    lazy_add[(node << 1) + 1] =
        lazy_add[(node << 1) + 1] * lazy_mul[node] + lazy_add[node];
    int m = (s + e) / 2;
    tree[2 * node] *= lazy_mul[node];
    tree[2 * node + 1] *= lazy_mul[node];
    tree[2 * node] += (m - s + 1) * lazy_add[node];
    tree[2 * node + 1] += (e - m) * lazy_add[node];
    tree[node] = tree[2 * node] + tree[2 * node + 1];
  }
  lazy_mul[node] = 1;
  lazy_add[node] = 0;
}
void build(int node, int s, int e) {
  lazy_mul[node] = 1;
  if (s == e)
    tree[node] = ary[s];
  else {
    int m = (s + e) / 2;
    build((node << 1), s, m);
    build((node << 1) + 1, m + 1, e);
    tree[node] = tree[(node << 1)] + tree[(node << 1) + 1];
  }
}
double query(int node, int s, int e, int l, int r) {
  if (s > r || e < l) return 0;
  if (s >= l && e <= r) return tree[node];
  push(node, s, e);
  int m = (s + e) >> 1;
  double x = query((node << 1), s, m, l, r);
  double y = query((node << 1) + 1, m + 1, e, l, r);
  return x + y;
}
void update(int node, int s, int e, int l, int r, double mul, double add) {
  if (s > e || s > r || e < l) return;
  if (s >= l && e <= r) {
    tree[node] = tree[node] * mul + (e - s + 1) * add;
    lazy_mul[node] *= mul;
    lazy_add[node] = lazy_add[node] * mul + add;
    return;
  }
  push(node, s, e);
  int m = (s + e) >> 1;
  update(node << 1, s, m, l, r, mul, add);
  update((node << 1) + 1, m + 1, e, l, r, mul, add);
  tree[node] = tree[node << 1] + tree[(node << 1) + 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> ary[i];
  build(1, 0, n - 1);
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      double l1, l2, r1, r2, x, y;
      cin >> l1 >> r1 >> l2 >> r2;
      l1--, l2--, r1--, r2--;
      double q1 = query(1, 0, n - 1, l1, r1);
      double q2 = query(1, 0, n - 1, l2, r2);
      x = r1 - l1 + 1;
      y = r2 - l2 + 1;
      update(1, 0, n - 1, l1, r1, (x - 1) / x, q2 / (x * y));
      update(1, 0, n - 1, l2, r2, (y - 1) / y, q1 / (x * y));
    } else {
      double l, r;
      cin >> l >> r;
      l--, r--;
      cout << setprecision(7) << fixed << query(1, 0, n - 1, l, r) << "\n";
    }
  }
  return 0;
}