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;
template <typename T1, typename T2>
bool Max(T1& a, T2 b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T1, typename T2>
bool Min(T1& a, T2 b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T, typename AD>
struct ST {
  vector<T> st;
  vector<AD> lazy;
  int n, h;
  static T combine(T a, T b) { return a + b; }
  static T defaultT() { return 0; }
  void add(int i, int sz, AD change) {
    st[i] = change.apply(st[i], sz);
    if (sz > 1) {
      lazy[i].combine(change);
    }
  }
  int hsb_loc(int n) { return 31 - __builtin_clz(n); }
  void init(int _n) {
    n = _n;
    h = hsb_loc(n) + 1;
    st.assign(n << 1, defaultT());
    lazy.assign(n, AD());
  }
  void init(T* a, int _n) {
    init(_n);
    for (int i = 0; i < n; i++) {
      st[n + i] = a[i];
    }
    for (int i = n - 1; i > 0; i--) {
      st[i] = combine(st[i << 1], st[i << 1 | 1]);
    }
  }
  void push(int p) {
    for (int s = h; s > 0; --s) {
      int i = p >> s;
      if (lazy[i]) {
        add(i << 1, 1 << (s - 1), lazy[i]);
        add(i << 1 | 1, 1 << (s - 1), lazy[i]);
        lazy[i] = AD();
      }
    }
  }
  T query(int l, int r) {
    l += n;
    r += n;
    push(l);
    push(r);
    T ret = defaultT();
    for (; l <= r; l >>= 1, r >>= 1) {
      if (l & 1) ret = combine(ret, st[l++]);
      if (!(r & 1)) ret = combine(ret, st[r--]);
    }
    return ret;
  }
  void update(int l, int r, AD update) {
    l += n;
    r += n;
    push(l);
    push(r);
    int l0 = l, r0 = r;
    for (int k = 1; l <= r; l >>= 1, r >>= 1, k <<= 1) {
      if (l & 1) add(l++, k, update);
      if (!(r & 1)) add(r--, k, update);
    }
    while (!(l0 & 1)) l0 >>= 1;
    for (int i = l0; i > 1; i >>= 1) {
      st[i >> 1] = combine(st[i], st[i ^ 1]);
    }
    while (r0 & 1) r0 >>= 1;
    for (int i = r0; i > 1; i >>= 1) {
      st[i >> 1] = combine(st[i], st[i ^ 1]);
    }
  }
};
struct upd {
  double m, a;
  upd() : m(1), a(0) {}
  upd(double _m, double _a) : m(_m), a(_a) {}
  double apply(double num, int sz) { return num * m + a * sz; }
  double combine(upd rhs) {
    m *= rhs.m;
    a = a * rhs.m + rhs.a;
  }
  operator bool() { return m != 1 || a != 0; }
};
const int N = 2000001, inf = 1e9;
double a[N];
void solve() {
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < (n); ++i) cin >> a[i];
  ST<double, upd> st;
  st.init(a, n);
  while (q--) {
    int t, l1, r1, l2, r2;
    cin >> t >> l1 >> r1;
    l1--, r1--;
    if (t == 1) {
      cin >> l2 >> r2;
      l2--, r2--;
      double n1 = r1 - l1 + 1, n2 = r2 - l2 + 1;
      double exp1 = st.query(l1, r1) / n1;
      double exp2 = st.query(l2, r2) / n2;
      st.update(l1, r1, upd((n1 - 1) / n1, exp2 / n1));
      st.update(l2, r2, upd((n2 - 1) / n2, exp1 / n2));
    } else {
      double ans = st.query(l1, r1);
      cout << ans << '\n';
    }
  }
}
void init() {}
void gen() {}
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(20);
  if (argc > 1 && strcmp(argv[1], "gen") == 0) {
    gen();
    return 0;
  }
  init();
  int TC = 1;
  for (int TI = 1; TI <= (TC); ++TI) {
    solve();
  }
  return 0;
}