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 maxn = 1e5 + 7;
struct Seg {
  double rate;
  double add;
  double sum;
  inline void Up(double r, double a, int l) {
    sum = sum * r + a * l;
    add = add * r + a;
    rate *= r;
  }
} st[maxn << 2];
inline void Pushup(int rt) {
  st[rt].sum = st[(rt << 1)].sum + st[(rt << 1 | 1)].sum;
}
inline void Build(int L, int R, int rt) {
  st[rt].rate = 1;
  if (L == R) {
    cin >> st[rt].sum;
    return;
  }
  int Mid = L + R >> 1;
  ;
  Build(L, Mid, (rt << 1));
  Build(Mid + 1, R, (rt << 1 | 1));
  Pushup(rt);
}
ostream& operator<<(ostream& out, const Seg& A) {
  out << "rate :" << A.rate << "add:" << A.add << "sum:" << A.sum << endl;
  return out;
}
inline void Modify(int L, int R, int l, int r, double a, double b, int rt) {
  if (l <= L && r >= R) {
    st[rt].Up(b, a, (R - L + 1));
    return;
  }
  int Mid = L + R >> 1;
  ;
  double& _r = st[rt].rate;
  double& _add = st[rt].add;
  st[(rt << 1)].Up(_r, _add, (Mid - L + 1));
  st[(rt << 1 | 1)].Up(_r, _add, (R - Mid));
  _r = 1, _add = 0;
  if (l <= Mid) Modify(L, Mid, l, r, a, b, (rt << 1));
  if (r > Mid) Modify(Mid + 1, R, l, r, a, b, (rt << 1 | 1));
  Pushup(rt);
}
inline double Query(int L, int R, int l, int r, int rt) {
  if (l <= L && r >= R) {
    return st[rt].sum;
  }
  int Mid = L + R >> 1;
  ;
  double& _r = st[rt].rate;
  double& _add = st[rt].add;
  st[(rt << 1)].Up(_r, _add, (Mid - L + 1));
  st[(rt << 1 | 1)].Up(_r, _add, (R - Mid));
  _r = 1, _add = 0;
  double re = 0;
  if (l <= Mid) re = Query(L, Mid, l, r, (rt << 1));
  if (r > Mid) re += Query(Mid + 1, R, l, r, (rt << 1 | 1));
  return re;
}
int main() {
  cout << fixed;
  int n, q;
  cin >> n >> q;
  Build(1, n, 1);
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      double s1 = Query(1, n, l2, r2, 1) / (r2 - l2 + 1.0) / (r1 - l1 + 1.0);
      double s2 = Query(1, n, l1, r1, 1) / (r2 - l2 + 1.0) / (r1 - l1 + 1.0);
      Modify(1, n, l1, r1, s1, 1.0 * (r1 - l1) / (r1 - l1 + 1.0), 1);
      Modify(1, n, l2, r2, s2, 1.0 * (r2 - l2) / (r2 - l2 + 1.0), 1);
    } else {
      int l, r;
      cin >> l >> r;
      cout << Query(1, n, l, r, 1) << endl;
      ;
    }
  }
  return 0;
}