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>
const int maxN = 1e5 + 7;
struct Node {
  int l, r;
  double sum, lazy, mul;
} tree[maxN << 2];
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
void init(int now, int l, int r) {
  tree[now].l = l;
  tree[now].r = r;
  tree[now].mul = 1;
  tree[now].lazy = 0;
  return;
}
void updata(int now) {
  tree[now].sum = tree[now << 1].sum + tree[now << 1 | 1].sum;
  return;
}
void build(int now, int l, int r) {
  init(now, l, r);
  if (l == r) {
    tree[now].sum = read();
    return;
  }
  int mid = (l + r) >> 1;
  build(now << 1, l, mid);
  build(now << 1 | 1, mid + 1, r);
  updata(now);
  return;
}
void work(int now, double lazy, double mul) {
  tree[now].lazy = tree[now].lazy * mul + lazy;
  tree[now].mul = tree[now].mul * mul;
  tree[now].sum = tree[now].sum * mul + lazy * (tree[now].r - tree[now].l + 1);
  return;
}
void pushdown(int now) {
  work(now << 1, tree[now].lazy, tree[now].mul);
  work(now << 1 | 1, tree[now].lazy, tree[now].mul);
  tree[now].lazy = 0;
  tree[now].mul = 1;
  return;
}
double query(int now, int L, int R) {
  if (tree[now].l >= L && tree[now].r <= R) return tree[now].sum;
  int mid = (tree[now].l + tree[now].r) >> 1;
  double sum = 0;
  if (tree[now].lazy || tree[now].mul != 1) pushdown(now);
  if (mid >= L) sum += query(now << 1, L, R);
  if (mid < R) sum += query(now << 1 | 1, L, R);
  return sum;
}
void modify(int now, int L, int R, double lazy, double mul) {
  if (tree[now].l >= L && tree[now].r <= R) {
    work(now, lazy, mul);
    return;
  }
  if (tree[now].lazy || tree[now].mul != 1) pushdown(now);
  int mid = (tree[now].l + tree[now].r) >> 1;
  if (mid >= L) modify(now << 1, L, R, lazy, mul);
  if (mid < R) modify(now << 1 | 1, L, R, lazy, mul);
  updata(now);
  return;
}
int main() {
  int n, m;
  n = read();
  m = read();
  build(1, 1, n);
  int opt, l1, r1, l2, r2;
  while (m--) {
    opt = read() - 1;
    l1 = read();
    r1 = read();
    if (!opt) {
      l2 = read();
      r2 = read();
      int L_1, L_2;
      double S_1, S_2;
      L_1 = r1 - l1 + 1;
      L_2 = r2 - l2 + 1;
      S_1 = query(1, l1, r1);
      S_2 = query(1, l2, r2);
      modify(1, l1, r1, 1.0 / L_1 * S_2 / L_2, 1.0 * (L_1 - 1) / L_1);
      modify(1, l2, r2, 1.0 / L_2 * S_1 / L_1, 1.0 * (L_2 - 1) / L_2);
    } else
      printf("%.7lf\n", query(1, l1, r1));
  }
  return 0;
}