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 MAXN = 100005;
int n, q, pot = 1;
double a, b;
struct node {
  double val, a, b;
  bool prop;
  node() { val = 0, a = 1, b = 0, prop = 0; }
  node(double _val, double _a, double _b) {
    val = _val;
    a = _a;
    b = _b;
  }
} t[4 * MAXN];
void propagate(int cvor, int low, int high) {
  t[cvor].prop = 0;
  t[cvor].val = t[cvor].a * t[cvor].val + t[cvor].b * (high - low + 1);
  if (cvor >= pot) {
    t[cvor].a = 1;
    t[cvor].b = 0;
    return;
  }
  if (t[cvor].a != 1) {
    t[cvor * 2].a *= t[cvor].a;
    t[cvor * 2].b *= t[cvor].a;
    t[cvor * 2 + 1].a *= t[cvor].a;
    t[cvor * 2 + 1].b *= t[cvor].a;
    t[cvor].a = 1;
  }
  if (t[cvor].b != 0) {
    t[cvor * 2].b += t[cvor].b;
    t[cvor * 2 + 1].b += t[cvor].b;
    t[cvor].b = 0;
  }
  t[cvor * 2].prop = t[cvor * 2 + 1].prop = 1;
}
void update(int cvor, int from, int to, int low, int high) {
  if (t[cvor].prop) propagate(cvor, low, high);
  if (from <= low && high <= to) {
    t[cvor].a = a;
    t[cvor].b = b;
    propagate(cvor, low, high);
    return;
  }
  if (to < low || high < from) return;
  update(cvor * 2, from, to, low, (low + high) / 2);
  update(cvor * 2 + 1, from, to, (low + high) / 2 + 1, high);
  t[cvor].val = t[cvor * 2].val + t[cvor * 2 + 1].val;
}
double upit(int cvor, int from, int to, int low, int high) {
  if (t[cvor].prop) propagate(cvor, low, high);
  if (from <= low && high <= to) return t[cvor].val;
  if (to < low || high < from) return 0;
  return upit(cvor * 2, from, to, low, (low + high) / 2) +
         upit(cvor * 2 + 1, from, to, (low + high) / 2 + 1, high);
}
int main() {
  cin >> n >> q;
  while (pot < n) pot *= 2;
  for (int i = 0; i < n; i++) {
    scanf("%lf", &t[i + pot]);
  }
  for (int i = pot - 1; i > 0; i--) {
    t[i].val = t[i * 2].val + t[i * 2 + 1].val;
  }
  for (int i = 0; i < q; i++) {
    int tip, l1, r1, l2, r2;
    scanf("%d", &tip);
    if (tip == 1) {
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double s1 = upit(1, l1 - 1, r1 - 1, 0, pot - 1);
      double s2 = upit(1, l2 - 1, r2 - 1, 0, pot - 1);
      a = ((double)r1 - l1) / (r1 - l1 + 1);
      b = s2 / (r1 - l1 + 1) / (r2 - l2 + 1);
      update(1, l1 - 1, r1 - 1, 0, pot - 1);
      a = ((double)r2 - l2) / (r2 - l2 + 1);
      b = s1 / (r1 - l1 + 1) / (r2 - l2 + 1);
      update(1, l2 - 1, r2 - 1, 0, pot - 1);
    } else {
      scanf("%d%d", &l1, &r1);
      printf("%.9f\n", upit(1, l1 - 1, r1 - 1, 0, pot - 1));
    }
  }
  return 0;
}