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>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
long long inf = 1e9 + 7;
long long INF = 1e18;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
vector<vector<long long> > g;
vector<long long> order;
vector<long long> r, l;
struct inc_tree {
  long long n;
  vector<double> t;
  vector<double> mod;
  vector<double> mult;
  inc_tree(vector<double> a = vector<double>(0)) {
    n = 1;
    while (n < a.size()) n *= 2;
    t.assign(n * 4 + 10, 0);
    mod.assign(n * 4 + 10, 0);
    mult.assign(n * 4 + 10, 1);
    a.resize(n);
    build(a, 1, 0, n - 1);
  }
  void build(vector<double> &a, long long v, long long tl, long long tr) {
    if (tl == tr)
      t[v] = a[tl];
    else {
      long long tm = (tl + tr) / 2;
      build(a, v * 2, tl, tm);
      build(a, v * 2 + 1, tm + 1, tr);
      t[v] = t[v * 2 + 1] + t[v * 2];
    }
  }
  void push(long long v, long long tl, long long tr) {
    if (mod[v] == 0 && mult[v] == 1) return;
    if (tl == tr) {
      t[v] *= mult[v];
      t[v] += mod[v];
      mod[v] = 0;
      mult[v] = 1;
      return;
    }
    t[v] *= mult[v];
    t[v] += mod[v] * (tr - tl + 1);
    mult[v * 2] *= mult[v];
    mult[v * 2 + 1] *= mult[v];
    mod[v * 2] *= mult[v];
    mod[v * 2 + 1] *= mult[v];
    mod[v * 2] += mod[v];
    mod[v * 2 + 1] += mod[v];
    mod[v] = 0;
    mult[v] = 1;
  }
  double sum(long long l, long long r, long long v = 1, long long tl = 0,
             long long tr = -1) {
    if (l > r) return 0;
    if (tr == -1) tr = n - 1;
    if (mod[v] != 0 || mult[v] != 1) push(v, tl, tr);
    if (tl == tr) return t[v];
    if (tl == l && tr == r) return t[v];
    long long tm = (tl + tr) / 2;
    return sum(l, min(r, tm), v * 2, tl, tm) +
           sum(max(l, tm + 1), r, v * 2 + 1, tm + 1, tr);
  }
  void update(long long l, long long r, double y, double x, long long v = 1,
              long long tl = 0, long long tr = -1) {
    if (l > r) return;
    if (tr == -1) tr = n - 1;
    if (mod[v] != 0 || mult[v] != 1) push(v, tl, tr);
    if (tl == tr) {
      mult[v] *= y;
      mod[v] *= y;
      mod[v] += x;
      return;
    }
    if (l == tl && r == tr) {
      mult[v] *= y;
      mod[v] *= y;
      mod[v] += x;
      return;
    }
    push(v, tl, tr);
    long long tm = (tl + tr) / 2;
    update(l, min(r, tm), y, x, v * 2, tl, tm);
    update(max(l, tm + 1), r, y, x, v * 2 + 1, tm + 1, tr);
    push(v * 2, tl, tm);
    push(v * 2 + 1, tm + 1, tr);
    t[v] = t[v * 2] + t[v * 2 + 1];
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(6);
  int n, q;
  cin >> n >> q;
  vector<double> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  inc_tree T(a);
  for (int i = 0; i < q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      l1--;
      r1--;
      l2--;
      r2--;
      double l = r1 - l1 + 1;
      double r = r2 - l2 + 1;
      double a1 = (l - 1) / l;
      double b1 = T.sum(l2, r2) / (r * l);
      double a2 = (r - 1) / r;
      double b2 = T.sum(l1, r1) / (r * l);
      T.update(l1, r1, a1, b1);
      T.update(l2, r2, a2, b2);
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << T.sum(l, r) << '\n';
    }
  }
  return 0;
}