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 <class T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) {
  _W(x.first);
  putchar(' ');
  _W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
template <class T, class... U>
void DEBUG(const T &head, const U &...tail) {}
int MOD = 1e9 + 7;
void ADD(int64_t &x, int64_t v) {
  x = (x + v) % MOD;
  if (x < 0) x += MOD;
}
const int SIZE = 4e5 + 10;
int a[SIZE], num[SIZE];
double s[SIZE], _tmp[SIZE][2];
void push(int id) {
  if (_tmp[id] == 0) return;
  _tmp[id * 2][0] *= _tmp[id][0];
  _tmp[id * 2][1] = _tmp[id * 2][1] * _tmp[id][0] + _tmp[id][1];
  s[id * 2] = _tmp[id][0] * s[id * 2] + _tmp[id][1] * num[id * 2];
  _tmp[id * 2 + 1][0] *= _tmp[id][0];
  _tmp[id * 2 + 1][1] = _tmp[id * 2 + 1][1] * _tmp[id][0] + _tmp[id][1];
  s[id * 2 + 1] = _tmp[id][0] * s[id * 2 + 1] + _tmp[id][1] * num[id * 2 + 1];
  _tmp[id][0] = 1;
  _tmp[id][1] = 0;
}
void pull(int id) { s[id] = s[id * 2] + s[id * 2 + 1]; }
void init(int L, int R, int id) {
  _tmp[id][0] = 1;
  _tmp[id][1] = 0;
  if (L == R) {
    s[id] = a[L];
    num[id] = 1;
    return;
  }
  int mm = (L + R) / 2;
  init(L, mm, id * 2);
  init(mm + 1, R, id * 2 + 1);
  num[id] = num[id * 2] + num[id * 2 + 1];
  pull(id);
}
double qq(int L, int R, int id, int ll, int rr) {
  if (R < ll || L > rr) return 0;
  if (L >= ll && R <= rr) return s[id];
  int mm = (L + R) / 2;
  push(id);
  return qq(L, mm, id * 2, ll, rr) + qq(mm + 1, R, id * 2 + 1, ll, rr);
}
double c1, c2;
void ins(int L, int R, int id, int ll, int rr) {
  if (R < ll || L > rr) return;
  if (L >= ll && R <= rr) {
    s[id] = s[id] * c1 + c2 * num[id];
    _tmp[id][0] *= c1;
    _tmp[id][1] = _tmp[id][1] * c1 + c2;
    return;
  }
  int mm = (L + R) / 2;
  push(id);
  ins(L, mm, id * 2, ll, rr);
  ins(mm + 1, R, id * 2 + 1, ll, rr);
  pull(id);
}
int main() {
  int n, q;
  R(n, q);
  for (int i = (1); i <= (n); ++i) R(a[i]);
  init(1, n, 1);
  while (q--) {
    int ty;
    R(ty);
    if (ty == 1) {
      int l1, r1, l2, r2;
      R(l1, r1, l2, r2);
      int n1 = r1 - l1 + 1;
      int n2 = r2 - l2 + 1;
      double v1 = qq(1, n, 1, l1, r1);
      double v2 = qq(1, n, 1, l2, r2);
      c1 = (n1 - 1.) / n1;
      c2 = v2 / ((int64_t)n1 * n2);
      ins(1, n, 1, l1, r1);
      c1 = (n2 - 1.) / n2;
      c2 = v1 / ((int64_t)n1 * n2);
      ins(1, n, 1, l2, r2);
    } else {
      int l, r;
      R(l, r);
      printf("%.9f\n", qq(1, n, 1, l, r));
    }
  }
  return 0;
}