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 N = 1e5 + 9;
int n, m;
struct tree {
  double tot, ad, mul;
} tr[N << 2];
long long a[N];
void build(int now, int l, int r) {
  tr[now].ad = 0, tr[now].mul = 1;
  if (l == r) {
    tr[now].tot = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build(now * 2, l, mid);
  build(now * 2 + 1, mid + 1, r);
  tr[now].tot = tr[now * 2].tot + tr[now * 2 + 1].tot;
}
void pushdown(int now, int l, int r) {
  int mid = (l + r) / 2;
  tr[now * 2].tot = tr[now].mul * tr[now * 2].tot + tr[now].ad * (mid - l + 1);
  tr[now * 2 + 1].tot =
      tr[now].mul * tr[now * 2 + 1].tot + tr[now].ad * (r - mid);
  tr[2 * now].mul *= tr[now].mul, tr[2 * now + 1].mul *= tr[now].mul;
  tr[2 * now].ad = tr[2 * now].ad * tr[now].mul + tr[now].ad;
  tr[2 * now + 1].ad = tr[2 * now + 1].ad * tr[now].mul + tr[now].ad;
  tr[now].mul = 1, tr[now].ad = 0;
}
void pls(int now, int lx, int rx, int l, int r, double d) {
  if (r < lx || l > rx) return;
  if (l <= lx && r >= rx) {
    tr[now].ad += d, tr[now].tot += (rx - lx + 1) * d;
    return;
  }
  pushdown(now, lx, rx);
  int mid = (lx + rx) / 2;
  pls(now * 2, lx, mid, l, r, d);
  pls(now * 2 + 1, mid + 1, rx, l, r, d);
  tr[now].tot = tr[now * 2].tot + tr[now * 2 + 1].tot;
}
void mulm(int now, int lx, int rx, int l, int r, double d) {
  if (r < lx || l > rx) return;
  if (l <= lx && r >= rx) {
    tr[now].mul *= d, tr[now].tot *= d, tr[now].ad *= d;
    return;
  }
  pushdown(now, lx, rx);
  int mid = (lx + rx) / 2;
  mulm(now * 2, lx, mid, l, r, d);
  mulm(now * 2 + 1, mid + 1, rx, l, r, d);
  tr[now].tot = tr[now * 2].tot + tr[now * 2 + 1].tot;
}
double query(int now, int lx, int rx, int l, int r) {
  if (r < lx || l > rx) return 0;
  if (rx <= r && lx >= l) return tr[now].tot;
  pushdown(now, lx, rx);
  int mid = (lx + rx) / 2;
  double ans =
      query(now * 2, lx, mid, l, r) + query(now * 2 + 1, mid + 1, rx, l, r);
  return ans;
}
signed main(int argc, char** argv) {
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      double x = query(1, 1, n, l1, r1), y = query(1, 1, n, l2, r2),
             ave1 = x / (double)(r1 - l1 + 1), ave2 = y / (double)(r2 - l2 + 1);
      mulm(1, 1, n, l1, r1, 1.0 - 1 / (double)(r1 - l1 + 1));
      mulm(1, 1, n, l2, r2, 1.0 - 1 / (double)(r2 - l2 + 1));
      pls(1, 1, n, l1, r1, ave2 / (double)(r1 - l1 + 1)),
          pls(1, 1, n, l2, r2, ave1 / (double)(r2 - l2 + 1));
    } else {
      int l, r;
      cin >> l >> r;
      printf("%.10lf\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}