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 mod = 1000000007;
double it[400009], add[400009], mul[400009], a[100009];
long long n, m;
void init(long long k, long long l, long long r) {
  mul[k] = 1;
  add[k] = 0;
  if (l == r) {
    it[k] = a[l];
    return;
  }
  long long mid = (l + r) / 2;
  init(k * 2, l, mid);
  init(k * 2 + 1, mid + 1, r);
  it[k] = it[k * 2] + it[k * 2 + 1];
}
void dolazy(long long k, long long l, long long r) {
  it[k] *= mul[k];
  it[k] += add[k] * (r - l + 1);
  if (l < r) {
    mul[k * 2] *= mul[k];
    mul[k * 2 + 1] *= mul[k];
    add[k * 2] = add[k * 2] * mul[k] + add[k];
    add[k * 2 + 1] = add[k * 2 + 1] * mul[k] + add[k];
  }
  mul[k] = 1;
  add[k] = 0;
}
void upd(long long k, long long l, long long r, long long L, long long R,
         double mu, double ad) {
  dolazy(k, l, r);
  if (r < L || r < l || R < l) return;
  long long mid = (l + r) / 2;
  if (L <= l && r <= R) {
    it[k] *= mu;
    it[k] += ad * (r - l + 1);
    if (l < r) {
      mul[k * 2] *= mu;
      mul[k * 2 + 1] *= mu;
      add[k * 2] = add[k * 2] * mu + ad;
      add[k * 2 + 1] = add[k * 2 + 1] * mu + ad;
    }
    return;
  }
  upd(k * 2, l, mid, L, R, mu, ad);
  upd(k * 2 + 1, mid + 1, r, L, R, mu, ad);
  it[k] = it[k * 2] + it[k * 2 + 1];
}
double getsum(long long k, long long l, long long r, long long L, long long R) {
  dolazy(k, l, r);
  if (r < L || r < l || R < l) return 0.0;
  if (L <= l && r <= R) return it[k];
  long long mid = (l + r) / 2;
  return getsum(k * 2, l, mid, L, R) + getsum(k * 2 + 1, mid + 1, r, L, R);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(7);
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  init(1, 1, n);
  while (m--) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      double s1 = getsum(1, 1, n, l1, r1);
      double s2 = getsum(1, 1, n, l2, r2);
      double v1 = r1 - l1 + 1;
      double v2 = r2 - l2 + 1;
      upd(1, 1, n, l1, r1, (v1 - 1) / v1, s2 / v2 / v1);
      upd(1, 1, n, l2, r2, (v2 - 1) / v2, s1 / v1 / v2);
    } else {
      long long x, y;
      cin >> x >> y;
      cout << getsum(1, 1, n, x, y) << '\n';
    }
  }
}