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;
struct ST {
  double key, mul, add;
} a[400003];
int n, m;
int s[100003];
int ls(int x) { return x << 1; }
int rs(int x) { return x << 1 | 1; }
void push_up(int x) { a[x].key = a[ls(x)].key + a[rs(x)].key; }
void build(int x, int l, int r) {
  a[x].mul = 1;
  a[x].add = 0;
  if (l == r) {
    a[x].key = s[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(ls(x), l, mid);
  build(rs(x), mid + 1, r);
  push_up(x);
}
void push_down(int x, int l, int r) {
  int mid = (l + r) >> 1;
  a[ls(x)].key = a[ls(x)].key * a[x].mul + a[x].add * (mid - l + 1);
  a[rs(x)].key = a[rs(x)].key * a[x].mul + a[x].add * (r - mid);
  a[ls(x)].mul *= a[x].mul;
  a[rs(x)].mul *= a[x].mul;
  a[ls(x)].add = a[ls(x)].add * a[x].mul + a[x].add;
  a[rs(x)].add = a[rs(x)].add * a[x].mul + a[x].add;
  a[x].mul = 1;
  a[x].add = 0;
}
void modify1(int x, int l, int r, int nl, int nr, double k) {
  if (nl <= l && nr >= r) {
    a[x].key *= k;
    a[x].mul *= k;
    a[x].add *= k;
    return;
  }
  push_down(x, l, r);
  int mid = (l + r) >> 1;
  if (nl <= mid) modify1(ls(x), l, mid, nl, nr, k);
  if (nr > mid) modify1(rs(x), mid + 1, r, nl, nr, k);
  push_up(x);
}
void modify2(int x, int l, int r, int nl, int nr, double k) {
  if (nl <= l && nr >= r) {
    a[x].key += (r - l + 1) * k;
    a[x].add += k;
    return;
  }
  push_down(x, l, r);
  int mid = (l + r) >> 1;
  if (nl <= mid) modify2(ls(x), l, mid, nl, nr, k);
  if (nr > mid) modify2(rs(x), mid + 1, r, nl, nr, k);
  push_up(x);
}
double query(int x, int l, int r, int nl, int nr) {
  if (nl <= l && nr >= r) return a[x].key;
  push_down(x, l, r);
  int mid = (l + r) >> 1;
  double res = 0;
  if (nl <= mid) res = query(ls(x), l, mid, nl, nr);
  if (nr > mid) res += query(rs(x), mid + 1, r, nl, nr);
  return res;
}
void modify(int l, int r, int l1, int r1) {
  int len = r - l + 1, len1 = r1 - l1 + 1;
  double x = query(1, 1, n, l, r) / len, x1 = query(1, 1, n, l1, r1) / len1;
  modify1(1, 1, n, l, r, (double)(len - 1) / len);
  modify1(1, 1, n, l1, r1, (double)(len1 - 1) / len1);
  modify2(1, 1, n, l, r, x1 / len);
  modify2(1, 1, n, l1, r1, x / len1);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &s[i]);
  build(1, 1, n);
  while (m--) {
    int opt, l, r;
    scanf("%d%d%d", &opt, &l, &r);
    if (opt == 1) {
      int l1, r1;
      scanf("%d%d", &l1, &r1);
      modify(l, r, l1, r1);
    } else
      printf("%.7lf\n", query(1, 1, n, l, r));
  }
  return 0;
}