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 int NMAX = 100 * 1000;
struct Node {
  double exp, a, b;
};
Node aint[4 * NMAX + 2];
int N, Q;
double v[NMAX + 2];
pair<double, double> Composition(double a, double b, double x, double y) {
  return {a * x, a * y + b};
}
void Propagate(Node &a, Node &x, Node &y, int lg) {
  pair<double, double> st = Composition(a.a, a.b, x.a, x.b);
  pair<double, double> dr = Composition(a.a, a.b, y.a, y.b);
  x.a = st.first;
  x.b = st.second;
  y.a = dr.first;
  y.b = dr.second;
  a.exp = a.a * a.exp + a.b * lg;
  a.a = 1;
  a.b = 0;
}
void ClearLazy(int nod, int st, int dr) {
  if (st != dr)
    Propagate(aint[nod], aint[2 * nod], aint[2 * nod + 1], dr - st + 1);
  else
    aint[nod].exp = aint[nod].exp * aint[nod].a + aint[nod].b, aint[nod].a = 1,
    aint[nod].b = 0;
}
void build(int nod, int st, int dr) {
  if (st == dr) {
    aint[nod].a = 1;
    aint[nod].b = 0;
    aint[nod].exp = v[st];
    return;
  }
  int mid = (st + dr) / 2;
  build(2 * nod, st, mid);
  build(2 * nod + 1, mid + 1, dr);
  aint[nod].exp = aint[2 * nod].exp + aint[2 * nod + 1].exp;
  aint[nod].a = 1;
  aint[nod].b = 0;
}
void Update(int nod, int st, int dr, int x, int y, double a, double b) {
  if (y < st || x > dr) return;
  ClearLazy(nod, st, dr);
  if (x <= st && dr <= y) {
    aint[nod].a = a;
    aint[nod].b = b;
    return;
  }
  int md = (st + dr) / 2;
  Update(2 * nod, st, md, x, y, a, b);
  Update(2 * nod + 1, md + 1, dr, x, y, a, b);
  ClearLazy(2 * nod, st, md);
  ClearLazy(2 * nod + 1, md + 1, dr);
  aint[nod].exp = aint[2 * nod].exp + aint[2 * nod + 1].exp;
}
double Query(int nod, int st, int dr, int x, int y) {
  if (y < st || x > dr) return 0.0;
  ClearLazy(nod, st, dr);
  if (x <= st && dr <= y) return aint[nod].exp;
  int md = (st + dr) / 2;
  return Query(2 * nod, st, md, x, y) + Query(2 * nod + 1, md + 1, dr, x, y);
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  cin >> N >> Q;
  for (int i = 1; i <= N; ++i) cin >> v[i];
  build(1, 1, N);
  for (int i = 1; i <= Q; ++i) {
    int op;
    cin >> op;
    if (op == 1) {
      int a, b, x, y;
      cin >> a >> b >> x >> y;
      double e1 = Query(1, 1, N, a, b) / (b - a + 1);
      double e2 = Query(1, 1, N, x, y) / (y - x + 1);
      Update(1, 1, N, a, b, 1.0 * (b - a) / (b - a + 1), e2 / (b - a + 1));
      Update(1, 1, N, x, y, 1.0 * (y - x) / (y - x + 1), e1 / (y - x + 1));
    } else {
      int x, y;
      cin >> x >> y;
      cout << Query(1, 1, N, x, y) << '\n';
    }
  }
  return 0;
}