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 = 1e5 + 10;
int n, q, a[maxn];
struct node {
  double sum;
  double lazyMult, lazySum;
} seg[4 * maxn];
void build(int idx, int i, int j) {
  seg[idx].lazyMult = 1.0;
  seg[idx].lazySum = 0;
  if (i == j) {
    seg[idx].sum = a[i];
    return;
  }
  int mid = (i + j) >> 1;
  int left = idx << 1;
  int right = left | 1;
  build(left, i, mid);
  build(right, mid + 1, j);
  seg[idx].sum = seg[left].sum + seg[right].sum;
}
void refresh(int idx, int i, int j) {
  int left = idx << 1;
  int right = left | 1;
  double mult = seg[idx].lazyMult;
  double sum = seg[idx].lazySum;
  if (i != j) {
    seg[left].lazyMult *= mult;
    seg[right].lazyMult *= mult;
    seg[left].lazySum = seg[left].lazySum * mult + sum;
    seg[right].lazySum = seg[right].lazySum * mult + sum;
  }
  seg[idx].sum *= mult;
  seg[idx].sum += sum * (j - i + 1);
  seg[idx].lazySum = 0;
  seg[idx].lazyMult = 1.0;
}
void update(int idx, int i, int j, int l, int r, double mult, double add) {
  refresh(idx, i, j);
  if (i > j || j < l || r < i) return;
  if (l <= i && j <= r) {
    seg[idx].lazyMult *= mult;
    seg[idx].lazySum = seg[idx].lazySum * mult + add;
    refresh(idx, i, j);
    return;
  }
  int mid = (i + j) >> 1;
  int left = idx << 1;
  int right = left | 1;
  update(left, i, mid, l, r, mult, add);
  update(right, mid + 1, j, l, r, mult, add);
  seg[idx].sum = seg[left].sum + seg[right].sum;
}
double query(int idx, int i, int j, int l, int r) {
  refresh(idx, i, j);
  if (i > j || j < l || r < i) return 0LL;
  if (l <= i && j <= r) return seg[idx].sum;
  int mid = (i + j) >> 1;
  int left = idx << 1;
  int right = left | 1;
  return query(left, i, mid, l, r) + query(right, mid + 1, j, l, r);
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  build(1, 1, n);
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      double sz1 = r1 - l1 + 1;
      double sz2 = r2 - l2 + 1;
      double sum1 = query(1, 1, n, l1, r1) / sz1;
      double sum2 = query(1, 1, n, l2, r2) / sz2;
      update(1, 1, n, l1, r1, (sz1 - 1) / sz1, sum2 / sz1);
      update(1, 1, n, l2, r2, (sz2 - 1) / sz2, sum1 / sz2);
    } else {
      int l, r;
      cin >> l >> r;
      cout << fixed << setprecision(10) << query(1, 1, n, l, r) << '\n';
    }
  }
  return 0;
}