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 p;
double tmp[200005];
struct node {
  double dat, tag1, tag2;
  int l, r;
};
struct seg {
  node d[800005];
  void pushup(int k) { d[k].dat = d[k << 1].dat + d[k << 1 | 1].dat; }
  void build(int k, int l, int r) {
    d[k].l = l;
    d[k].r = r;
    d[k].tag1 = 0;
    d[k].tag2 = 1;
    if (l == r) {
      d[k].dat = tmp[l];
      return;
    }
    int mid = (l + r) >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
    pushup(k);
  }
  void add(int k, double x) {
    double len = d[k].r - d[k].l + 1;
    d[k].dat = (d[k].dat + x * len);
    d[k].tag1 = (d[k].tag1 + x);
  }
  void mul(int k, double x) {
    d[k].dat = d[k].dat * x;
    d[k].tag1 = d[k].tag1 * x;
    d[k].tag2 = d[k].tag2 * x;
  }
  void pushdown(int k) {
    if (fabs(d[k].tag2 - 1) > 1e-10) {
      mul(k << 1, d[k].tag2);
      mul(k << 1 | 1, d[k].tag2);
      d[k].tag2 = 1;
    }
    if (fabs(d[k].tag1) > 1e-10) {
      add(k << 1, d[k].tag1);
      add(k << 1 | 1, d[k].tag1);
      d[k].tag1 = 0;
    }
  }
  void add(int k, int l, int r, double x) {
    if (l <= d[k].l && d[k].r <= r) {
      add(k, x);
      return;
    }
    pushdown(k);
    int mid = (d[k].l + d[k].r) >> 1;
    if (l <= mid) add(k << 1, l, r, x);
    if (r > mid) add(k << 1 | 1, l, r, x);
    pushup(k);
  }
  void mul(int k, int l, int r, double x) {
    if (l <= d[k].l && d[k].r <= r) {
      mul(k, x);
      return;
    }
    pushdown(k);
    int mid = (d[k].l + d[k].r) >> 1;
    if (l <= mid) mul(k << 1, l, r, x);
    if (r > mid) mul(k << 1 | 1, l, r, x);
    pushup(k);
  }
  double query(int k, int l, int r) {
    if (l <= d[k].l && d[k].r <= r) {
      return d[k].dat;
    }
    pushdown(k);
    double sum = 0;
    int mid = (d[k].l + d[k].r) >> 1;
    if (l <= mid) sum = (sum + query(k << 1, l, r));
    if (r > mid) sum = (sum + query(k << 1 | 1, l, r));
    return sum;
  }
} Seg;
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lf", &tmp[i]);
  Seg.build(1, 1, n);
  while (m--) {
    int opt, l1, r1, l2, r2;
    long long x;
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double d1 = Seg.query(1, l1, r1);
      double d2 = Seg.query(1, l2, r2);
      Seg.mul(1, l1, r1, double(r1 - l1) / double(r1 - l1 + 1));
      Seg.mul(1, l2, r2, double(r2 - l2) / double(r2 - l2 + 1));
      Seg.add(1, l1, r1, d2 / double(r1 - l1 + 1) / double(r2 - l2 + 1));
      Seg.add(1, l2, r2, d1 / double(r2 - l2 + 1) / double(r1 - l1 + 1));
    }
    if (opt == 2) {
      scanf("%d%d", &l1, &r1);
      printf("%.8lf\n", Seg.query(1, l1, r1));
    }
  }
  return 0;
}