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 INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 7;
const double pi = acos(-1.0);
const int v2 = (mod + 1) / 2;
int n, q, a[maxn];
struct {
  int l, r;
  double sum, k, lazy;
} b[4 * maxn];
void push_up(int now) { b[now].sum = b[now << 1].sum + b[now << 1 | 1].sum; }
void push_down(int now) {
  double sum = b[now].sum, k = b[now].k, lazy = b[now].lazy;
  if (lazy != 0 || k != 1) {
    b[now].k = 1;
    b[now].lazy = 0;
    b[now << 1].k *= k;
    b[now << 1].lazy = b[now << 1].lazy * k + lazy;
    int l = b[now << 1].l, r = b[now << 1].r;
    b[now << 1].sum = b[now << 1].sum * k + (r - l + 1) * lazy;
    b[now << 1 | 1].k *= k;
    b[now << 1 | 1].lazy = b[now << 1 | 1].lazy * k + lazy;
    l = b[now << 1 | 1].l, r = b[now << 1 | 1].r;
    b[now << 1 | 1].sum = b[now << 1 | 1].sum * k + (r - l + 1) * lazy;
  }
}
void build(int now, int l, int r) {
  b[now].l = l, b[now].r = r;
  b[now].k = 1, b[now].lazy = 0;
  if (l == r) {
    b[now].sum = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(now << 1, l, mid);
  build(now << 1 | 1, mid + 1, r);
  push_up(now);
}
void update(int now, int l, int r, double k, double lazy) {
  int L = b[now].l, R = b[now].r;
  if (l <= L && R <= r) {
    b[now].sum = b[now].sum * k + (R - L + 1) * lazy;
    b[now].k *= k;
    b[now].lazy = b[now].lazy * k + lazy;
    return;
  }
  int mid = (L + R) >> 1;
  push_down(now);
  if (mid >= l) update(now << 1, l, min(r, mid), k, lazy);
  if (mid < r) update(now << 1 | 1, max(l, mid + 1), r, k, lazy);
  push_up(now);
}
double query(int now, int l, int r) {
  int L = b[now].l, R = b[now].r;
  if (l <= L && R <= r) {
    return b[now].sum;
  }
  double res = 0;
  int mid = (L + R) >> 1;
  push_down(now);
  if (mid >= l) res += query(now << 1, l, min(r, mid));
  if (mid < r) res += query(now << 1 | 1, max(l, mid + 1), r);
  return res;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  build(1, 1, n);
  for (int i = 0; i < q; i++) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l1, r1, l2, r2;
      scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
      int num = r1 - l1 + 1, cnt = r2 - l2 + 1;
      double sum2 = query(1, l2, r2), sum1 = query(1, l1, r1);
      double k = 1.0 * (num - 1) / num, c = 1.0 / num / cnt * sum2;
      update(1, l1, r1, k, c);
      k = 1.0 * (cnt - 1) / cnt, c = 1.0 / cnt / num * sum1;
      update(1, l2, r2, k, c);
    } else {
      int l1, r1;
      scanf("%d %d", &l1, &r1);
      printf("%f\n", query(1, l1, r1));
    }
  }
}