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 MAX = 200010;
const long long MOD = (int)1e9 + 7;
const int INF = 1e9;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const long double EPS = 1e-8;
struct ST {
  long long id = 0;
  int size;
  vector<double> add, mul;
  vector<double> vec;
  double merge(double a, double b) { return a + b; }
  void init(int n) {
    size = 1;
    while (size < n) size *= 2;
    vec.assign(2 * size, 0);
    add.assign(2 * size, 0);
    mul.assign(2 * size, 0);
  }
  void propagate(int x, int lx, int rx) {
    if (abs(mul[x] - 1.0) > EPS) {
      vec[x] *= mul[x];
      if (rx - lx > 1) {
        add[2 * x + 1] *= mul[x];
        add[2 * x + 2] *= mul[x];
        mul[2 * x + 1] *= mul[x];
        mul[2 * x + 2] *= mul[x];
      }
      mul[x] = 1;
    }
    if (abs(add[x]) > EPS) {
      vec[x] += (double)(rx - lx) * add[x];
      if (rx - lx > 1) {
        add[2 * x + 1] += add[x];
        add[2 * x + 2] += add[x];
      }
      add[x] = 0;
    }
  }
  void update(int l, int r, double v, int type, int x, int lx, int rx) {
    int m = (rx + lx) / 2;
    propagate(x, lx, rx);
    if (r <= lx or l >= rx) return;
    if (l <= lx && rx <= r) {
      if (type == 1) {
        add[x] = v;
        propagate(x, lx, rx);
      }
      if (type == 2) {
        mul[x] = v;
        propagate(x, lx, rx);
      }
      return;
    }
    update(l, r, v, type, 2 * x + 1, lx, m);
    update(l, r, v, type, 2 * x + 2, m, rx);
    vec[x] = merge(vec[2 * x + 1], vec[2 * x + 2]);
  }
  void update(int l, int r, double v, int type) {
    update(l, r + 1, v, type, 0, 0, size);
  }
  double query(int l, int r, int x, int lx, int rx) {
    propagate(x, lx, rx);
    if (r <= lx or l >= rx) return 0;
    if (l <= lx && rx <= r) {
      return vec[x];
    }
    int m = (lx + rx) / 2;
    double s1 = query(l, r, 2 * x + 1, lx, m);
    double s2 = query(l, r, 2 * x + 2, m, rx);
    double res = merge(s1, s2);
    return res;
  }
  double query(int l, int r) { return query(l, r + 1, 0, 0, size); }
};
int32_t main() {
  int n, q;
  cin >> n >> q;
  vector<double> v;
  for (int i = 0; i < (int)n; i++) {
    int val;
    cin >> val;
    v.push_back((double)val);
  }
  ST st = ST();
  st.init(n + 1);
  for (int i = 0; i < (int)n; i++) {
    st.update(i, i, v[i], 1);
  }
  while (q--) {
    int tipo, l, r, l2, r2;
    cin >> tipo;
    if (tipo == 1) {
      cin >> l >> r >> l2 >> r2;
      l--;
      r--;
      l2--;
      r2--;
      int n1 = r - l + 1;
      int n2 = r2 - l2 + 1;
      double val1 = st.query(l, r);
      val1 /= (double)(n1);
      double val2 = st.query(l2, r2);
      val2 /= (double)(n2);
      st.update(l, r, (double)(n1 - 1.0) / (double)n1, 2);
      st.update(l2, r2, (double)(n2 - 1.0) / (double)n2, 2);
      st.update(l, r, val2 / (double)n1, 1);
      st.update(l2, r2, val1 / (double)n2, 1);
    } else {
      cin >> l >> r;
      l--;
      r--;
      cout << fixed << setprecision(8) << st.query(l, r) << '\n';
    }
  }
  return 0;
}