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 mod = 1e9 + 7;
const int INF = 2147483647;
struct node {
  node *l, *r;
  double a, b, sum;
  node() {
    l = r = NULL;
    a = 1, b = 0;
  }
};
int arr[100005];
void build(node *o, int l, int r) {
  if (l == r)
    o->sum = arr[l];
  else {
    int m = (l + r) / 2;
    o->l = new node(), build(o->l, l, m);
    o->r = new node(), build(o->r, m + 1, r);
    o->sum = (o->l->sum) + (o->r->sum);
  }
}
void push(node *o, int l, int r) {
  if (l < r) {
    int m = (l + r) / 2;
    o->l->sum = (o->a) * (o->l->sum) + (o->b) * (m - l + 1);
    o->l->a = (o->a) * (o->l->a);
    o->l->b = (o->a) * (o->l->b) + (o->b);
    o->r->sum = (o->a) * (o->r->sum) + (o->b) * (r - m);
    o->r->a = (o->a) * (o->r->a);
    o->r->b = (o->a) * (o->r->b) + (o->b);
  }
  o->a = 1, o->b = 0;
}
void modify(node *o, int l, int r, int ml, int mr, double x, double y) {
  if (ml <= l && r <= mr)
    o->sum = x * (o->sum) + y * (r - l + 1), o->a = x * (o->a),
    o->b = x * (o->b) + y;
  else {
    push(o, l, r);
    int m = (l + r) / 2;
    if (ml <= m) modify(o->l, l, m, ml, mr, x, y);
    if (mr > m) modify(o->r, m + 1, r, ml, mr, x, y);
    o->sum = (o->l->sum) + (o->r->sum);
  }
}
double qry(node *o, int l, int r, int ql, int qr) {
  if (qr < l || ql > r) return 0;
  if (ql <= l && r <= qr) return o->sum;
  push(o, l, r);
  int m = (l + r) / 2;
  return qry(o->l, l, m, ql, qr) + qry(o->r, m + 1, r, ql, qr);
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  node *root = new node();
  build(root, 1, n);
  while (q--) {
    int op, l1, r1;
    scanf("%d%d%d", &op, &l1, &r1);
    if (op == 1) {
      int l2, r2;
      scanf("%d%d", &l2, &r2);
      double s1 = qry(root, 1, n, l1, r1);
      double s2 = qry(root, 1, n, l2, r2);
      modify(root, 1, n, l1, r1, (double)(r1 - l1) / (r1 - l1 + 1),
             s2 / (r1 - l1 + 1) / (r2 - l2 + 1));
      modify(root, 1, n, l2, r2, (double)(r2 - l2) / (r2 - l2 + 1),
             s1 / (r1 - l1 + 1) / (r2 - l2 + 1));
    } else {
      printf("%lf\n", qry(root, 1, n, l1, r1));
    }
  }
  return 0;
}