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;
struct item {
  item() {}
  item(int v, int len) : len(len) {
    x = (double)v * len;
    a = b = push = 0;
  }
  double x, a, b;
  int len;
  bool push;
};
struct segtree {
  segtree(vector<int> a) : a(a), n(a.size()) {
    num.resize(n << 2);
    build(1, 0, n - 1);
  }
  void build(int p, int l, int r) {
    num[p] = item(a[l], r - l + 1);
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(p + p, l, mid);
    build(p + p + 1, mid + 1, r);
    combine(p, l, r);
  }
  void inject(int p, double c, double d) {
    double a = num[p].a;
    double b = num[p].b;
    int len = num[p].len;
    num[p].a = -(a * c - a - c);
    num[p].b = ((1 - c) * a * b + c * d) / num[p].a;
  }
  void relax(int p, int l, int r) {
    if (num[p].push) {
      if (l < r) {
        inject(p + p, num[p].a, num[p].b);
        inject(p + p + 1, num[p].a, num[p].b);
        num[p + p].push = 1;
        num[p + p + 1].push = 1;
      }
      num[p].x = (1 - num[p].a) * num[p].x + num[p].a * num[p].b * num[p].len;
      num[p].a = num[p].b = 0;
      num[p].push = 0;
    }
  }
  void combine(int p, int l, int r) {
    num[p].x = num[p + p].x + num[p + p + 1].x;
  }
  void update(int l, int r, double c, double d) {
    update(1, 0, n - 1, l, r, c, d);
  }
  void update(int p, int l, int r, int ll, int rr, double c, double d) {
    relax(p, l, r);
    if (ll <= l && r <= rr) {
      inject(p, c, d);
      num[p].push = 1;
      relax(p, l, r);
      return;
    }
    if (r < ll || rr < l) return;
    int mid = (l + r) >> 1;
    update(p + p, l, mid, ll, rr, c, d);
    update(p + p + 1, mid + 1, r, ll, rr, c, d);
    combine(p, l, r);
  }
  double find(int l, int r) { return find(1, 0, n - 1, l, r); }
  double find(int p, int l, int r, int ll, int rr) {
    relax(p, l, r);
    if (ll <= l && r <= rr) return num[p].x;
    if (r < ll || rr < l) return 0;
    int mid = (l + r) >> 1;
    double ret =
        find(p + p, l, mid, ll, rr) + find(p + p + 1, mid + 1, r, ll, rr);
    combine(p, l, r);
    return ret;
  }
  vector<item> num;
  vector<int> a;
  int n;
};
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  vector<int> a(n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  segtree seg(a);
  while (q--) {
    int t, l, r, l1, r1, l2, r2;
    scanf("%d", &t);
    if (t == 2) {
      scanf("%d %d", &l, &r);
      l--;
      r--;
      printf("%.15f\n", seg.find(l, r));
    } else {
      scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
      l1--;
      r1--;
      l2--;
      r2--;
      double lef = seg.find(l1, r1) / (r1 - l1 + 1);
      double rig = seg.find(l2, r2) / (r2 - l2 + 1);
      seg.update(l1, r1, 1. / (r1 - l1 + 1), rig);
      seg.update(l2, r2, 1. / (r2 - l2 + 1), lef);
    }
  }
  return 0;
}