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 int a[100005];
long double t[400005], sum0, n0;
pair<long double, long double> lazy[400005];
void f(long long int v, long double x, long double y) {
  lazy[v].first *= x;
  lazy[v].second *= x;
  lazy[v].second += y;
}
void push(long long int v, long long int tl, long long int tr) {
  long double &x = lazy[v].first;
  long double &y = lazy[v].second;
  f(2 * v, x, y);
  f(2 * v + 1, x, y);
  long long int tm = (tl + tr) / 2;
  t[2 * v] = t[2 * v] * x + y * (tm - tl + 1);
  t[2 * v + 1] = t[2 * v + 1] * x + y * (tr - tm);
  x = 1, y = 0;
}
void build(long long int v, long long int tl, long long int tr) {
  if (tl == tr) {
    t[v] = a[tl];
    return;
  }
  long long int tm = (tl + tr) / 2;
  build(2 * v, tl, tm);
  build(2 * v + 1, tm + 1, tr);
  t[v] = t[2 * v] + t[2 * v + 1];
}
void update(long long int v, long long int tl, long long int tr,
            long long int l, long long int r) {
  if (l > r) return;
  if (l == tl && r == tr) {
    t[v] = t[v] * n0 + (tr - tl + 1) * sum0;
    f(v, n0, sum0);
    return;
  }
  push(v, tl, tr);
  long long int tm = (tl + tr) / 2;
  update(2 * v, tl, tm, l, min(r, tm));
  update(2 * v + 1, tm + 1, tr, max(l, tm + 1), r);
  t[v] = t[2 * v] + t[2 * v + 1];
}
long double sum(long long int v, long long int tl, long long int tr,
                long long int l, long long int r) {
  if (l > r) return 0;
  if (l == tl && r == tr) return t[v];
  push(v, tl, tr);
  long long int tm = (tl + tr) / 2;
  return sum(2 * v, tl, tm, l, min(r, tm)) +
         sum(2 * v + 1, tm + 1, tr, max(l, tm + 1), r);
}
void solve() {
  long long int n, q;
  cin >> n >> q;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  for (long long int qc = 0; qc < q; qc++) {
    long long int op;
    cin >> op;
    if (op == 1) {
      long long int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      long double s1 = sum(1, 1, n, l1, r1), s2 = sum(1, 1, n, l2, r2);
      sum0 = s2 / (r2 - l2 + 1), n0 = r1 - l1 + 1;
      sum0 /= n0;
      n0 = (n0 - 1) / n0;
      update(1, 1, n, l1, r1);
      sum0 = s1 / (r1 - l1 + 1), n0 = r2 - l2 + 1;
      sum0 /= n0;
      n0 = (n0 - 1) / n0;
      update(1, 1, n, l2, r2);
    } else {
      long long int l, r;
      cin >> l >> r;
      cout << setprecision(100) << sum(1, 1, n, l, r) << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  clock_t z = clock();
  for (long long int i = 0; i < 400005; i++) lazy[i].first = 1;
  long long int qc = 1;
  for (long long int i = 1; i <= qc; i++) {
    solve();
  }
  fprintf(stderr, "Total Time:%.4Lf\n",
          (long double)(clock() - z) / CLOCKS_PER_SEC),
      fflush(stderr);
}