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 maxn = 1e6 + 10;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const long long INF = 1e15;
template <class T>
struct SegTree {
  struct node {
    T val, mul, add;
    node() : val(0), mul(1), add(0){};
  };
  vector<node> tr;
  SegTree(int n) { tr.resize(4 * n); }
  void build(int v, int l, int r, vector<T>& a) {
    if (l == r) {
      tr[v].val = a[l];
      return;
    }
    int mid = (l + r) / 2;
    build(v * 2, l, mid, a);
    build(v * 2 + 1, mid + 1, r, a);
    tr[v].val = tr[2 * v].val + tr[2 * v + 1].val;
  }
  void upd(int v, int l, int r, T add, T mul) {
    tr[v].mul *= mul;
    tr[v].add *= mul;
    tr[v].add += add;
    tr[v].val = tr[v].val * mul + add * (r - l + 1);
  }
  void push(int v, int l, int r) {
    int mid = (l + r) / 2;
    upd(2 * v, l, mid, tr[v].add, tr[v].mul);
    upd(2 * v + 1, mid + 1, r, tr[v].add, tr[v].mul);
    tr[v].mul = 1;
    tr[v].add = 0;
  }
  void update(int v, int l, int r, int x, int y, T add, T mul) {
    if (x == l and y == r) {
      upd(v, l, r, add, mul);
      return;
    }
    push(v, l, r);
    int mid = (l + r) / 2;
    if (x <= mid) update(2 * v, l, mid, x, min(y, mid), add, mul);
    if (y > mid) update(2 * v + 1, mid + 1, r, max(x, mid + 1), y, add, mul);
    tr[v].val = tr[2 * v].val + tr[2 * v + 1].val;
  }
  T query(int v, int l, int r, int x, int y) {
    if (x == l and y == r) return tr[v].val;
    push(v, l, r);
    T ans = T(0);
    int mid = (l + r) / 2;
    if (x <= mid) ans += query(2 * v, l, mid, x, min(y, mid));
    if (y > mid) ans += query(2 * v + 1, mid + 1, r, max(x, mid + 1), y);
    return ans;
  }
};
int main() {
  int n, q;
  scanf("%d", &(n));
  scanf("%d", &(q));
  vector<long double> a(n, 0);
  for (int i = 0; i < n; ++i) scanf("%Lf", &(a[i]));
  SegTree<long double> st(n);
  st.build(1, 0, n - 1, a);
  for (int i = 0; i < q; ++i) {
    int id;
    scanf("%d", &(id));
    if (id == 1) {
      int l1, r1, l2, r2;
      scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
      --l1, --r1, --l2, --r2;
      int len1 = r1 - l1 + 1, len2 = r2 - l2 + 1;
      long double mul1 = 1.0 * (len1 - 1) / len1,
                  mul2 = 1.0 * (len2 - 1) / len2;
      long double add1 = st.query(1, 0, n - 1, l2, r2) / len2 / len1;
      long double add2 = st.query(1, 0, n - 1, l1, r1) / len1 / len2;
      st.update(1, 0, n - 1, l1, r1, add1, mul1);
      st.update(1, 0, n - 1, l2, r2, add2, mul2);
    } else {
      int l, r;
      scanf("%d", &(l));
      scanf("%d", &(r));
      --l, --r;
      printf("%.7Lf\n", st.query(1, 0, n - 1, l, r));
    }
  }
  return 0;
}