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;
class SegmentTree {
 public:
  const static int N = 1e6 + 10;
  constexpr static double eps = 1e-6;
  double arr[N], st[N << 2], mul[N << 2], add[N << 2];
  inline void Push(int now, int l, int r) {
    if (abs(mul[now] - 1.0) > eps) {
      for (int i : {(now << 1), ((now << 1) + 1)}) {
        st[i] *= mul[now];
        mul[i] *= mul[now];
        add[i] *= mul[now];
      }
      mul[now] = 1.0;
    }
    if (add[now] > 0.0) {
      st[(now << 1)] +=
          (double)(((l + r) >> 1) - l + 1) / (double)(r - l + 1) * add[now];
      add[(now << 1)] +=
          (double)(((l + r) >> 1) - l + 1) / (double)(r - l + 1) * add[now];
      st[((now << 1) + 1)] +=
          (double)(r - ((l + r) >> 1)) / (double)(r - l + 1) * add[now];
      add[((now << 1) + 1)] +=
          (double)(r - ((l + r) >> 1)) / (double)(r - l + 1) * add[now];
      add[now] = 0.0;
    }
  }
  inline void Pull(int now, int l, int r) {
    st[now] = st[(now << 1)] + st[((now << 1) + 1)];
  }
  void Init(int now, int l, int r) {
    if (now == 1) {
      fill(mul, mul + (N << 2), 1.0);
      fill(add, add + (N << 2), 0.0);
    }
    if (l == r) {
      st[now] = arr[l];
      return;
    }
    Init((now << 1), l, ((l + r) >> 1));
    Init(((now << 1) + 1), ((l + r) >> 1) + 1, r);
    Pull(now, l, r);
  }
  void Mul(int ql, int qr, double interval, int now, int l, int r) {
    if (ql > qr or ql > r or qr < l) return;
    Push(now, l, r);
    if (l == ql and r == qr) {
      double now_value = (interval - 1) / interval;
      st[now] *= now_value;
      mul[now] *= now_value;
      add[now] *= now_value;
      return;
    }
    if (qr <= ((l + r) >> 1))
      Mul(ql, qr, interval, (now << 1), l, ((l + r) >> 1));
    else if (((l + r) >> 1) < ql)
      Mul(ql, qr, interval, ((now << 1) + 1), ((l + r) >> 1) + 1, r);
    else {
      Mul(ql, ((l + r) >> 1), interval, (now << 1), l, ((l + r) >> 1));
      Mul(((l + r) >> 1) + 1, qr, interval, ((now << 1) + 1),
          ((l + r) >> 1) + 1, r);
    }
    Pull(now, l, r);
  }
  void Add(int ql, int qr, double interval, double value, int now, int l,
           int r) {
    if (ql > qr or ql > r or qr < l) return;
    Push(now, l, r);
    if (l == ql and r == qr) {
      double now_value = value * (double)(r - l + 1) / interval;
      st[now] += now_value;
      add[now] += now_value;
      return;
    }
    if (qr <= ((l + r) >> 1))
      Add(ql, qr, interval, value, (now << 1), l, ((l + r) >> 1));
    else if (((l + r) >> 1) < ql)
      Add(ql, qr, interval, value, ((now << 1) + 1), ((l + r) >> 1) + 1, r);
    else {
      Add(ql, ((l + r) >> 1), interval, value, (now << 1), l, ((l + r) >> 1));
      Add(((l + r) >> 1) + 1, qr, interval, value, ((now << 1) + 1),
          ((l + r) >> 1) + 1, r);
    }
    Pull(now, l, r);
  }
  double Query(int ql, int qr, int now, int l, int r) {
    Push(now, l, r);
    if (l == ql and r == qr) return st[now];
    if (qr <= ((l + r) >> 1))
      return Query(ql, qr, (now << 1), l, ((l + r) >> 1));
    else if (((l + r) >> 1) < ql)
      return Query(ql, qr, ((now << 1) + 1), ((l + r) >> 1) + 1, r);
    else {
      double left = Query(ql, ((l + r) >> 1), (now << 1), l, ((l + r) >> 1));
      double right = Query(((l + r) >> 1) + 1, qr, ((now << 1) + 1),
                           ((l + r) >> 1) + 1, r);
      return left + right;
    }
  }
};
const int N = 1e5 + 10;
int n, q, a[N];
SegmentTree st;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) st.arr[i] = a[i];
  st.Init(1, 1, n);
  cout << fixed << setprecision(12);
  for (int _ = 0; _ < q; ++_) {
    int type;
    cin >> type;
    if (type == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      double d1 = r1 - l1 + 1;
      double d2 = r2 - l2 + 1;
      double sum1 = st.Query(l1, r1, 1, 1, n);
      double sum2 = st.Query(l2, r2, 1, 1, n);
      st.Mul(l1, r1, d1, 1, 1, n);
      st.Add(l1, r1, d1, sum2 / d2, 1, 1, n);
      st.Mul(l2, r2, d2, 1, 1, n);
      st.Add(l2, r2, d2, sum1 / d1, 1, 1, n);
    } else {
      int l, r;
      cin >> l >> r;
      cout << st.Query(l, r, 1, 1, n) << '\n';
    }
  }
}