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 double eps = 1e-7;
struct segtree {
  struct node {
    double sum;
    double mul, add;
    node() {
      sum = 0;
      mul = 1;
      add = 0;
    }
    void apply_mul(int l, int r, double v) {
      mul *= v;
      add *= v;
      sum *= v;
    }
    void apply_add(int l, int r, double v) {
      add += v;
      sum += v * (r - l + 1);
    }
  };
  inline void push(int x, int l, int r) {
    int m = (l + r) >> 1;
    int z = x + ((m - l + 1) << 1);
    if (fabs(tree[x].mul - 1) > eps) {
      tree[x + 1].apply_mul(l, m, tree[x].mul);
      tree[z].apply_mul(m + 1, r, tree[x].mul);
      tree[x].mul = 1;
    }
    if (fabs(tree[x].add) > eps) {
      tree[x + 1].apply_add(l, m, tree[x].add);
      tree[z].apply_add(m + 1, r, tree[x].add);
      tree[x].add = 0;
    }
  }
  node unite(const node &a, const node &b) const {
    node res;
    res.sum = a.sum + b.sum;
    return res;
  }
  inline void pull(int x, int z) { tree[x] = unite(tree[x + 1], tree[z]); }
  int n;
  vector<node> tree;
  void build(int x, int l, int r, const vector<double> &v) {
    if (l == r) {
      tree[x].sum = v[l];
      return;
    }
    int m = (l + r) >> 1;
    int z = x + ((m - l + 1) << 1);
    build(x + 1, l, m, v);
    build(z, m + 1, r, v);
    pull(x, z);
  }
  node get(int x, int l, int r, int ll, int rr) {
    if (ll <= l && r <= rr) {
      return tree[x];
    }
    int y = (l + r) >> 1;
    int z = x + ((y - l + 1) << 1);
    push(x, l, r);
    node res;
    if (rr <= y) {
      res = get(x + 1, l, y, ll, rr);
    } else {
      if (ll > y) {
        res = get(z, y + 1, r, ll, rr);
      } else {
        res = unite(get(x + 1, l, y, ll, rr), get(z, y + 1, r, ll, rr));
      }
    }
    pull(x, z);
    return res;
  }
  void modify_mul(int x, int l, int r, int ll, int rr, double v) {
    if (ll <= l && r <= rr) {
      tree[x].apply_mul(l, r, v);
      return;
    }
    int m = (l + r) >> 1;
    int z = x + ((m - l + 1) << 1);
    push(x, l, r);
    if (ll <= m) {
      modify_mul(x + 1, l, m, ll, rr, v);
    }
    if (rr > m) {
      modify_mul(z, m + 1, r, ll, rr, v);
    }
    pull(x, z);
  }
  void modify_add(int x, int l, int r, int ll, int rr, double v) {
    if (ll <= l && r <= rr) {
      tree[x].apply_add(l, r, v);
      return;
    }
    int m = (l + r) >> 1;
    int z = x + ((m - l + 1) << 1);
    push(x, l, r);
    if (ll <= m) {
      modify_add(x + 1, l, m, ll, rr, v);
    }
    if (rr > m) {
      modify_add(z, m + 1, r, ll, rr, v);
    }
    pull(x, z);
  }
  segtree() {}
  template <typename M>
  segtree(const vector<M> &v) {
    n = v.size();
    assert(n > 0);
    tree.resize(2 * n - 1);
    build(0, 0, n - 1, v);
  }
  node get(int ll, int rr) {
    assert(0 <= ll && ll <= rr && rr <= n - 1);
    return get(0, 0, n - 1, ll, rr);
  }
  void modify_mul(int ll, int rr, double v) {
    assert(0 <= ll && ll <= rr && rr <= n - 1);
    modify_mul(0, 0, n - 1, ll, rr, v);
  }
  void modify_add(int ll, int rr, double v) {
    assert(0 <= ll && ll <= rr && rr <= n - 1);
    modify_add(0, 0, n - 1, ll, rr, v);
  }
};
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  vector<double> a(n);
  for (int i = 0; i < n; i++) {
    scanf("%lf", &a[i]);
  }
  segtree st(a);
  while (q--) {
    int tp, l1, r1, l2, r2, l, r;
    scanf("%d", &tp);
    if (tp == 1) {
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      l1--;
      r1--;
      l2--;
      r2--;
      int lena = r1 - l1 + 1;
      int lenb = r2 - l2 + 1;
      double suma = st.get(l1, r1).sum;
      double sumb = st.get(l2, r2).sum;
      st.modify_mul(l1, r1, 1.0 * (lena - 1) / lena);
      st.modify_add(l1, r1, sumb / lena / lenb);
      st.modify_mul(l2, r2, 1.0 * (lenb - 1) / lenb);
      st.modify_add(l2, r2, suma / lena / lenb);
    } else {
      scanf("%d%d", &l, &r);
      l--;
      r--;
      printf("%.8f\n", st.get(l, r).sum);
    }
  }
  return 0;
}