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;
long long int n, q;
long long int a[100005];
double tree[400005];
double lazy[2][400005];
void lazify(long long int i, long long int s, long long int e) {
  if (lazy[0][i] != 1 || lazy[1][i] != 0) {
    tree[i] = lazy[0][i] * tree[i] + lazy[1][i] * double(e - s + 1);
    if (s != e) {
      lazy[0][2 * i] = lazy[0][2 * i] * lazy[0][i];
      lazy[1][2 * i] = lazy[1][2 * i] * lazy[0][i] + lazy[1][i];
      lazy[0][2 * i + 1] = lazy[0][2 * i + 1] * lazy[0][i];
      lazy[1][2 * i + 1] = lazy[1][2 * i + 1] * lazy[0][i] + lazy[1][i];
    }
    lazy[0][i] = 1.00000;
    lazy[1][i] = 0.00000;
  }
}
double qur(long long int l, long long int r, long long int s, long long int e,
           long long int i) {
  if (s > r || e < l) return 0.00000;
  lazify(i, s, e);
  if (s >= l && e <= r) return tree[i];
  long long int mid = (s + e) >> 1;
  return (qur(l, r, s, mid, 2 * i) + qur(l, r, mid + 1, e, 2 * i + 1));
}
void upd(long long int l, long long int r, long long int s, long long int e,
         long long int i, double a, double b) {
  lazify(i, s, e);
  if (s > r || e < l) return;
  if (s >= l && e <= r) {
    lazy[0][i] = lazy[0][i] * a;
    lazy[1][i] = lazy[1][i] * a + b;
    lazify(i, s, e);
    return;
  }
  long long int mid = (s + e) >> 1;
  upd(l, r, s, mid, 2 * i, a, b);
  upd(l, r, mid + 1, e, 2 * i + 1, a, b);
  tree[i] = tree[2 * i] + tree[2 * i + 1];
}
void build(long long int l, long long int r, long long int i) {
  if (l == r) {
    tree[i] = double(a[l]);
    return;
  }
  long long int mid = (l + r) >> 1;
  build(l, mid, 2 * i);
  build(mid + 1, r, 2 * i + 1);
  tree[i] = tree[2 * i] + tree[2 * i + 1];
}
int main() {
  for (int i = 1; i <= 400000; i++) {
    lazy[0][i] = 1.00000;
    lazy[1][i] = 0.00000;
  }
  scanf("%lld", &n);
  scanf("%lld", &q);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  build(1, n, 1);
  for (int i = 1; i <= q; i++) {
    long long int c, l1, r1, l2, r2;
    scanf("%lld", &c);
    if (c == 1) {
      scanf("%lld", &l1);
      scanf("%lld", &r1);
      scanf("%lld", &l2);
      scanf("%lld", &r2);
      double x = qur(l2, r2, 1, n, 1);
      double y = qur(l1, r1, 1, n, 1);
      upd(l1, r1, 1, n, 1, double(r1 - l1) / double(r1 - l1 + 1),
          (x / double(r2 - l2 + 1)) * (1.00000 / double(r1 - l1 + 1)));
      upd(l2, r2, 1, n, 1, double(r2 - l2) / double(r2 - l2 + 1),
          (y / double(r1 - l1 + 1)) * (1.00000 / double(r2 - l2 + 1)));
    } else {
      scanf("%lld", &l1);
      scanf("%lld", &r1);
      double x = qur(l1, r1, 1, n, 1);
      printf("%0.6f\n", x);
    }
  }
  return 0;
}