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:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int inf = numeric_limits<int>::max() / 2;
const double eps = 1e-9;
const double pi = acosl(-1);
template <typename T>
inline bool mineq(T& a, T b) {
  return (a > b) ? (a = b, 1) : 0;
}
template <typename T>
inline bool maxeq(T& a, T b) {
  return (a < b) ? (a = b, 1) : 0;
}
inline void solve(), read();
const string file = "";
signed main() {
  if (file != "") {
    freopen((file + ".in").c_str(), "r", stdin);
    freopen((file + ".out").c_str(), "w", stdout);
  }
  read();
  solve();
  return 0;
}
struct vertex {
  double add = 0;
  double mult = 1;
  double sum = 0;
};
struct segment {
  int n;
  vector<vertex> t;
  segment() {}
  segment(vector<int>& a) {
    int LOG = log2l((int)((a).size())) + 1;
    n = 1ll << LOG;
    t.resize(4 * n);
    build(a);
  }
  void build(vector<int>& a) {
    for (int i = 0; i < (int)((a).size()); i++) {
      t[n + i].sum = a[i];
    }
    for (int i = n - 1; i >= 0; i--) {
      t[i].sum = t[2 * i].sum + t[2 * i + 1].sum;
    }
  }
  void push(int v, int tl, int tr) {
    t[v].sum = t[v].sum * t[v].mult + t[v].add * double(tr - tl + 1);
    t[2 * v].mult *= t[v].mult;
    t[2 * v + 1].mult *= t[v].mult;
    t[2 * v].add *= t[v].mult;
    t[2 * v + 1].add *= t[v].mult;
    t[2 * v].add += t[v].add;
    t[2 * v + 1].add += t[v].add;
    t[v].add = 0;
    t[v].mult = 1;
  }
  void upd(int v, int tl, int tr, int l, int r, double mult, double add) {
    push(v, tl, tr);
    if (tl > r || tr < l) {
      return;
    }
    if (l <= tl && tr <= r) {
      t[v].add += add;
      t[v].mult *= mult;
      push(v, tl, tr);
      return;
    }
    int tm = (tl + tr) / 2;
    upd(2 * v, tl, tm, l, r, mult, add);
    upd(2 * v + 1, tm + 1, tr, l, r, mult, add);
    t[v].sum = t[2 * v].sum + t[2 * v + 1].sum;
  }
  double get(int v, int tl, int tr, int l, int r) {
    push(v, tl, tr);
    if (tl > r || tr < l) {
      return 0;
    }
    if (l <= tl && tr <= r) {
      return t[v].sum;
    }
    int tm = (tl + tr) / 2;
    double v1 = get(2 * v, tl, tm, l, r);
    double v2 = get(2 * v + 1, tm + 1, tr, l, r);
    return v1 + v2;
  }
};
int n, q;
vector<int> a;
segment st;
inline void solve() {
  cout << fixed << setprecision(5);
  st = segment(a);
  for (int i = 0; i < q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      double l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      --l1, --r1, --l2, --r2;
      double a1 = st.get(1, 0, st.n - 1, l1, r1) / (r1 - l1 + 1);
      double a2 = st.get(1, 0, st.n - 1, l2, r2) / (r2 - l2 + 1);
      st.upd(1, 0, st.n - 1, l1, r1, (r1 - l1) / (r1 - l1 + 1),
             a2 / (r1 - l1 + 1));
      st.upd(1, 0, st.n - 1, l2, r2, (r2 - l2) / (r2 - l2 + 1),
             a1 / (r2 - l2 + 1));
    } else if (t == 2) {
      double l, r;
      cin >> l >> r;
      --l, --r;
      double ans = st.get(1, 0, st.n - 1, l, r);
      cout << ans << "\n";
    }
  }
}
inline void read() {
  cin >> n >> q;
  a.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
}