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;
inline int read() {
  int x;
  char c;
  int f = 1;
  while ((c = getchar()) != '-' && (c < '0' || c > '9'))
    ;
  if (c == '-') c = getchar(), f = -1;
  x = c ^ '0';
  while ((c = getchar()) >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ '0');
  return x * f;
}
inline long long readll() {
  long long x;
  char c;
  long long f = 1;
  while ((c = getchar()) != '-' && (c < '0' || c > '9'))
    ;
  if (c == '-') c = getchar(), f = -1;
  x = c ^ '0';
  while ((c = getchar()) >= '0' && c <= '9')
    x = (x << 1ll) + (x << 3ll) + (c ^ '0');
  return x * f;
}
const int maxn = 1e5 + 10;
namespace Seg {
double sum[maxn << 2], rev[maxn << 2], addv[maxn << 2];
void push_up(int x) { sum[x] = sum[x << 1] + sum[x << 1 | 1]; }
void push_down(int x, int L, int R) {
  if (rev[x] != 1) {
    sum[x << 1] *= rev[x];
    sum[x << 1 | 1] *= rev[x];
    rev[x << 1] *= rev[x];
    rev[x << 1 | 1] *= rev[x];
    addv[x << 1] *= rev[x];
    addv[x << 1 | 1] *= rev[x];
    rev[x] = 1;
  }
  if (addv[x]) {
    int Mid = (L + R) >> 1;
    sum[x << 1] += (Mid - L + 1) * addv[x];
    sum[x << 1 | 1] += (R - Mid) * addv[x];
    addv[x << 1] += addv[x];
    addv[x << 1 | 1] += addv[x];
    addv[x] = 0;
  }
}
void build_tree(int x, int L, int R) {
  rev[x] = 1;
  if (L == R) {
    sum[x] = read();
    return;
  }
  int Mid = (L + R) >> 1;
  build_tree(x << 1, L, Mid);
  build_tree(x << 1 | 1, Mid + 1, R);
  push_up(x);
}
double query(int x, int L, int R, int ql, int qr) {
  if (ql <= L && R <= qr) return sum[x];
  int Mid = (L + R) >> 1;
  push_down(x, L, R);
  double ss = 0;
  if (ql <= Mid) ss += query(x << 1, L, Mid, ql, qr);
  if (qr > Mid) ss += query(x << 1 | 1, Mid + 1, R, ql, qr);
  push_up(x);
  return ss;
}
void update(int x, int L, int R, int ql, int qr, double vc, double va) {
  if (ql <= L && R <= qr) {
    rev[x] *= vc;
    addv[x] *= vc;
    addv[x] += va;
    sum[x] *= vc;
    sum[x] += va * (R - L + 1);
    return;
  }
  int Mid = (L + R) >> 1;
  push_down(x, L, R);
  if (ql <= Mid) update(x << 1, L, Mid, ql, qr, vc, va);
  if (qr > Mid) update(x << 1 | 1, Mid + 1, R, ql, qr, vc, va);
  push_up(x);
}
}  // namespace Seg
int main() {
  int n = read(), m = read();
  Seg::build_tree(1, 1, n);
  while (m--) {
    int ty = read();
    if (ty == 1) {
      int l1 = read(), r1 = read(), l2 = read(), r2 = read();
      double sum1 = Seg::query(1, 1, n, l1, r1),
             sum2 = Seg::query(1, 1, n, l2, r2);
      double x1 = r1 - l1 + 1, x2 = r2 - l2 + 1;
      Seg::update(1, 1, n, l1, r1, (x1 - 1) / x1, sum2 / (x2 * x1));
      Seg::update(1, 1, n, l2, r2, (x2 - 1) / x2, sum1 / (x2 * x1));
    } else {
      int l = read(), r = read();
      printf("%lf\n", Seg::query(1, 1, n, l, r));
    }
  }
  return 0;
}