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 long long int N = 100005;
int n, q;
long long int a[N];
struct node {
  double val, lz0, lz1;
  node() {
    val = lz0 = 0;
    lz1 = 1;
  }
} st[4 * N];
void build(int id, int l, int r) {
  if (l == r) {
    st[id].val = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build(id * 2, l, mid);
  build(id * 2 + 1, mid + 1, r);
  st[id].val = st[id * 2].val + st[id * 2 + 1].val;
}
void down(int id, int l, int r) {
  if (l == r) return;
  double t = st[id * 2].val;
  int mid = (l + r) / 2;
  st[id * 2].val = st[id * 2].val * st[id].lz1 + st[id].lz0 * (mid - l + 1);
  st[id * 2 + 1].val = st[id * 2 + 1].val * st[id].lz1 + st[id].lz0 * (r - mid);
  st[id * 2].lz1 = st[id].lz1 * st[id * 2].lz1;
  st[id * 2 + 1].lz1 = st[id].lz1 * st[id * 2 + 1].lz1;
  st[id * 2].lz0 = st[id].lz1 * st[id * 2].lz0 + st[id].lz0;
  st[id * 2 + 1].lz0 = st[id].lz1 * st[id * 2 + 1].lz0 + st[id].lz0;
  st[id].lz1 = 1;
  st[id].lz0 = 0;
}
void update(int id, int l, int r, int u, int v, double a, double b) {
  if (v < l || r < u) return;
  down(id, l, r);
  if (u <= l && r <= v) {
    st[id].val = st[id].val * a + b * (r - l + 1);
    st[id].lz1 *= a;
    st[id].lz0 = st[id].lz0 * a + b;
    return;
  }
  int mid = (l + r) / 2;
  update(id * 2, l, mid, u, v, a, b);
  update(id * 2 + 1, mid + 1, r, u, v, a, b);
  st[id].val = st[id * 2].val + st[id * 2 + 1].val;
}
double get(int id, int l, int r, int u, int v) {
  if (v < l || r < u) return 0;
  down(id, l, r);
  if (u <= l && r <= v) {
    return st[id].val;
  }
  int mid = (l + r) / 2;
  return get(id * 2, l, mid, u, v) + get(id * 2 + 1, mid + 1, r, u, v);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%ld", &a[i]);
  build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    int c;
    scanf("%d", &c);
    if (c == 1) {
      int l1, r1, l2, r2;
      scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
      double v1 = (double)(r1 - l1 + 1), v2 = (double)(r2 - l2 + 1);
      double s1 = get(1, 1, n, l1, r1);
      double s2 = get(1, 1, n, l2, r2);
      update(1, 1, n, l1, r1, (double)(v1 - 1.0) / v1, s2 / 1.0 / v1 / v2);
      update(1, 1, n, l2, r2, (double)(v2 - 1.0) / v2, s1 / 1.0 / v1 / v2);
    } else {
      int l1, r1;
      scanf("%d %d", &l1, &r1);
      printf("%.9f \n", get(1, 1, n, l1, r1));
    }
  }
  return 0;
}