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, maxM = 500, MOD = 1e9 + 7;
const double EPS = 1e-10;
double T[maxN << 1];
pair<double, double> d[maxN];
int h, n, tams[maxN << 1];
void apply(int w, pair<double, double> val) {
  T[w] = T[w] * val.first + val.second * tams[w];
  if (w < n)
    d[w].first = val.first * d[w].first,
    d[w].second = val.first * d[w].second + val.second;
}
void pushDown(int w) {
  for (int i = h; i; i--) {
    int c = w >> i;
    if (d[c] != pair<double, double>(1, 0))
      apply(c << 1, d[c]), apply(c << 1 | 1, d[c]);
    d[c] = pair<double, double>(1, 0);
  }
}
void takeUp(int w) {
  while (w > 1)
    w >>= 1,
        T[w] = d[w].first * (T[w << 1] + T[w << 1 | 1]) + d[w].second * tams[w];
}
double query(int l, int r) {
  pushDown(l + n), pushDown(r + n - 1);
  double res = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) res += T[l++];
    if (r & 1) res += T[--r];
  }
  return res;
}
void update(int l, int r, pair<double, double> val) {
  int l0 = l += n, r0 = r += n;
  for (; l < r; l >>= 1, r >>= 1) {
    if (l & 1) apply(l++, val);
    if (r & 1) apply(--r, val);
  }
  takeUp(l0), takeUp(r0 - 1);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int q;
  cin >> n >> q;
  h = log2(n) + 1;
  for (int i = 0; i < n; i++) cin >> T[i + n], tams[i + n] = 1;
  for (int i = n - 1; i >= 0; i--)
    T[i] = T[i << 1] + T[i << 1 | 1], tams[i] = tams[i << 1] + tams[i << 1 | 1],
    d[i] = pair<double, double>(1, 0);
  while (q--) {
    int l1, l2, r1, r2, ty;
    cin >> ty >> l1 >> r1;
    l1--;
    if (ty == 1) {
      cin >> l2 >> r2;
      l2--;
      double tmp = query(l2, r2);
      update(l2, r2,
             pair<double, double>(1 - 1.0 / (r2 - l2),
                                  query(l1, r1) / (r1 - l1) / (r2 - l2)));
      update(l1, r1,
             pair<double, double>(1 - 1.0 / (r1 - l1),
                                  tmp / (r1 - l1) / (r2 - l2)));
    } else {
      cout << fixed << setprecision(9) << query(l1, r1) << '\n';
    }
  }
}