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 MAX = 1e5 + 9;
int n, a[MAX];
long double tree[(MAX << 2)], lazy[(MAX << 2)][2];
void Build(int low = 1, int high = n, int pos = 1) {
  lazy[pos][0] = 1.0;
  if (low == high) {
    tree[pos] = a[low];
    return;
  }
  int mid = ((low + high) >> 1);
  Build(low, mid, (pos << 1));
  Build(mid + 1, high, (pos << 1 | 1));
  tree[pos] = tree[(pos << 1)] + tree[(pos << 1 | 1)];
}
void Push(int low, int high, int pos) {
  tree[pos] *= lazy[pos][0];
  if (low != high) {
    lazy[(pos << 1)][0] *= lazy[pos][0];
    lazy[(pos << 1 | 1)][0] *= lazy[pos][0];
    lazy[(pos << 1)][1] *= lazy[pos][0];
    lazy[(pos << 1 | 1)][1] *= lazy[pos][0];
  }
  lazy[pos][0] = 1.0;
  tree[pos] += 1.0 * (high - low + 1) * lazy[pos][1];
  if (low != high) {
    lazy[(pos << 1)][1] += lazy[pos][1];
    lazy[(pos << 1 | 1)][1] += lazy[pos][1];
  }
  lazy[pos][1] = 0;
}
void PushAll(int low = 1, int high = 1, int pos = 1) {
  Push(low, high, pos);
  if (low == high) {
    return;
  }
  int mid = ((low + high) >> 1);
  PushAll(low, mid, (pos << 1));
  PushAll(mid + 1, high, (pos << 1 | 1));
  tree[pos] = tree[(pos << 1)] + tree[(pos << 1 | 1)];
}
void Update(bool id, int L, int R, long double val, int low = 1, int high = n,
            int pos = 1) {
  if (low == 1 && high == n) {
    PushAll();
  }
  Push(low, high, pos);
  if (L > high || R < low) {
    return;
  }
  if (L <= low && R >= high) {
    if (!id)
      lazy[pos][0] *= val;
    else
      lazy[pos][1] += val;
    Push(low, high, pos);
    return;
  }
  int mid = ((low + high) >> 1);
  Update(id, L, R, val, low, mid, (pos << 1));
  Update(id, L, R, val, mid + 1, high, (pos << 1 | 1));
  tree[pos] = tree[(pos << 1)] + tree[(pos << 1 | 1)];
}
long double Query(int L, int R, int low = 1, int high = n, int pos = 1) {
  if (low == 1 && high == n) {
    PushAll();
  }
  Push(low, high, pos);
  if (L > high || R < low) {
    return 0;
  }
  if (L <= low && R >= high) {
    return tree[pos];
  }
  int mid = ((low + high) >> 1);
  return Query(L, R, low, mid, (pos << 1)) +
         Query(L, R, mid + 1, high, (pos << 1 | 1));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int q;
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  Build();
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int L1, R1, L2, R2;
      cin >> L1 >> R1 >> L2 >> R2;
      long double sum1 = Query(L1, R1);
      long double sum2 = Query(L2, R2);
      Update(0, L1, R1, (long double)(R1 - L1) / (long double)(R1 - L1 + 1));
      Update(0, L2, R2, (long double)(R2 - L2) / (long double)(R2 - L2 + 1));
      Update(1, L1, R1,
             1.0 / (long double)(R1 - L1 + 1) * (long double)1.0 /
                 (long double)(R2 - L2 + 1) * sum2);
      Update(1, L2, R2,
             1.0 / (long double)(R2 - L2 + 1) * (long double)1.0 /
                 (long double)(R1 - L1 + 1) * sum1);
    } else {
      int L, R;
      cin >> L >> R;
      cout << setprecision(5) << Query(L, R) << "\n";
    }
  }
  return 0;
}