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 N = 1e5 + 5;
long long n, q, a[N];
double node[4 * N];
struct Lazy {
  double add, mul;
} lazy[4 * N];
void input() {
  scanf("%lld %lld", &n, &q);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
}
void init(long long i, long long l, long long r) {
  if (l > r) return;
  if (l == r) {
    node[i] = a[l] * 1.0;
    lazy[i].add = 0;
    lazy[i].mul = 1;
    return;
  }
  long long m = (l + r) >> 1;
  init(i << 1, l, m);
  init(i << 1 | 1, m + 1, r);
  node[i] = node[i << 1] + node[i << 1 | 1];
  lazy[i].add = 0;
  lazy[i].mul = 1;
}
void dolazy(long long i, long long l, long long r) {
  node[i] = node[i] * lazy[i].mul + lazy[i].add * (double)(r - l + 1);
  if (l != r) {
    lazy[i << 1].mul *= lazy[i].mul;
    lazy[i << 1 | 1].mul *= lazy[i].mul;
    lazy[i << 1].add = lazy[i << 1].add * lazy[i].mul + lazy[i].add;
    lazy[i << 1 | 1].add = lazy[i << 1 | 1].add * lazy[i].mul + lazy[i].add;
  }
  lazy[i].add = 0;
  lazy[i].mul = 1;
}
void update(long long i, long long l, long long r, long long a, long long b,
            double mul, double add) {
  dolazy(i, l, r);
  if (l > r || a > r || b < l) return;
  if (a <= l && r <= b) {
    node[i] = node[i] * mul + add * (double)(r - l + 1);
    if (l != r) {
      lazy[i << 1].add = lazy[i << 1].add * mul + add;
      lazy[i << 1 | 1].add = lazy[i << 1 | 1].add * mul + add;
      lazy[i << 1].mul *= mul;
      lazy[i << 1 | 1].mul *= mul;
    }
    return;
  }
  long long m = (l + r) >> 1;
  update(i << 1, l, m, a, b, mul, add);
  update(i << 1 | 1, m + 1, r, a, b, mul, add);
  node[i] = node[i << 1] + node[i << 1 | 1];
}
double get(long long i, long long l, long long r, long long a, long long b) {
  if (l > r || a > r || b < l) return 0;
  dolazy(i, l, r);
  if (a <= l && r <= b) return node[i];
  long long m = (l + r) >> 1;
  return get(i << 1, l, m, a, b) + get(i << 1 | 1, m + 1, r, a, b);
}
void solve() {
  while (q--) {
    long long type;
    scanf("%lld", &type);
    if (type == 1) {
      long long l1, r1, l2, r2;
      scanf("%lld %lld %lld %lld", &l1, &r1, &l2, &r2);
      double sum1 = get(1, 1, n, l1, r1);
      double sum2 = get(1, 1, n, l2, r2);
      sum1 = sum1 / (r1 - l1 + 1);
      sum2 = sum2 / (r2 - l2 + 1);
      update(1, 1, n, l1, r1, (double)(r1 - l1) / (double)(r1 - l1 + 1),
             sum2 / (double)(r1 - l1 + 1));
      update(1, 1, n, l2, r2, (double)(r2 - l2) / (double)(r2 - l2 + 1),
             sum1 / (double)(r2 - l2 + 1));
    } else {
      long long l, r;
      scanf("%lld %lld", &l, &r);
      printf("%.12f\n", get(1, 1, n, l, r));
    }
  }
}
signed main() {
  input();
  init(1, 1, n);
  solve();
  return 0;
}