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>
const char en = '\n';
using namespace std;
template <class T, class U>
ostream& operator<<(ostream& out, const pair<T, U>& par) {
  out << par.first << " " << par.second << en;
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  for (long long i = 0; i < (long long)(v.size()); i++) {
    if (i) out << " ";
    cout << v[i];
  }
  return out;
}
template <class ValueType>
class SegmentTree {
 private:
  int SIZE;
  int *Left, *Right;
  ValueType *value, *minimum, *maximum;
  ValueType *lazy_add, *lazy_mul_val;
  bool* lazy_mul;
  void clearVertex(int v) {
    if (lazy_mul[v]) {
      value[v] *= lazy_mul_val[v];
      lazy_mul[v] = false;
      if (Left[v] + 1 != Right[v]) {
        lazy_mul[2 * v] = lazy_mul[2 * v + 1] = true;
        lazy_mul_val[2 * v] *= lazy_mul_val[v];
        lazy_add[2 * v] *= lazy_mul_val[v];
        lazy_mul_val[2 * v + 1] *= lazy_mul_val[v];
        lazy_add[2 * v + 1] *= lazy_mul_val[v];
      }
      lazy_mul_val[v] = 1;
    }
    if (lazy_add[v]) {
      value[v] += lazy_add[v] * (Right[v] - Left[v]);
      if (Left[v] + 1 != Right[v]) {
        lazy_add[2 * v] += lazy_add[v];
        lazy_add[2 * v + 1] += lazy_add[v];
      }
      lazy_add[v] = 0;
    }
  }
  ValueType getSum(int v, int l, int r) {
    clearVertex(v);
    if (l == Left[v] && r == Right[v]) return value[v];
    int middle = (Left[v] + Right[v]) / 2;
    if (middle >= r) return getSum(2 * v, l, r);
    if (middle <= l) return getSum(2 * v + 1, l, r);
    return getSum(2 * v, l, middle) + getSum(2 * v + 1, middle, r);
  }
  void add(int v, int l, int r, ValueType val) {
    clearVertex(v);
    if (l == Left[v] && r == Right[v]) {
      lazy_add[v] = val;
      return;
    }
    int middle = (Left[v] + Right[v]) / 2;
    if (middle >= r)
      add(2 * v, l, r, val);
    else if (middle <= l)
      add(2 * v + 1, l, r, val);
    else {
      add(2 * v, l, middle, val);
      add(2 * v + 1, middle, r, val);
    }
    value[v] += (r - l) * val;
  }
  void mulValue(int v, int l, int r, ValueType val) {
    clearVertex(v);
    if (l == Left[v] && r == Right[v]) {
      lazy_mul_val[v] *= val;
      lazy_mul[v] = true;
      return;
    }
    int middle = (Left[v] + Right[v]) / 2;
    if (middle >= r)
      mulValue(2 * v, l, r, val);
    else if (middle <= l)
      mulValue(2 * v + 1, l, r, val);
    else {
      mulValue(2 * v, l, middle, val);
      mulValue(2 * v + 1, middle, r, val);
    }
    value[v] =
        (lazy_mul[2 * v] ? lazy_mul_val[2 * v] * value[2 * v] : value[2 * v]) +
        lazy_add[2 * v] * (Right[2 * v] - Left[2 * v]) +
        (lazy_mul[2 * v + 1] ? lazy_mul_val[2 * v + 1] * value[2 * v + 1]
                             : value[2 * v + 1]) +
        lazy_add[2 * v + 1] * (Right[2 * v + 1] - Left[2 * v + 1]);
  }

 public:
  void init(int len) {
    for (SIZE = 1; SIZE < len; SIZE *= 2)
      ;
    Left = (int*)malloc(sizeof(int) * 2 * SIZE);
    Right = (int*)malloc(sizeof(int) * 2 * SIZE);
    value = (ValueType*)malloc(sizeof(ValueType) * 2 * SIZE);
    lazy_add = (ValueType*)malloc(sizeof(ValueType) * 2 * SIZE);
    lazy_mul_val = (ValueType*)malloc(sizeof(ValueType) * 2 * SIZE);
    lazy_mul = (bool*)malloc(sizeof(bool) * 2 * SIZE);
    Left[1] = 0;
    Right[1] = SIZE;
    for (long long i = (long long)(1); i <= (long long)(SIZE - 1); i++) {
      Left[2 * i] = Left[i];
      Right[2 * i] = (Left[i] + Right[i]) / 2;
      Left[2 * i + 1] = (Left[i] + Right[i]) / 2;
      Right[2 * i + 1] = Right[i];
    }
    for (long long i = 0; i < (long long)(2 * SIZE); i++) {
      value[i] = lazy_add[i] = ValueType();
      lazy_mul[i] = false;
      lazy_mul_val[i] = 1;
    }
  }
  ValueType getSum(int l, int r) {
    if (l == r) return 0;
    return getSum(1, l, r);
  }
  void add(int l, int r, ValueType val) { add(1, l, r, val); }
  void mulValue(int l, int r, ValueType val) { mulValue(1, l, r, val); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  SegmentTree<long double> S;
  int n, q;
  cin >> n >> q;
  S.init(n + 1);
  vector<long double> V(n + 1, 0);
  for (long long i = 0; i < (long long)(n); i++) {
    int x;
    cin >> x;
    V[i + 1] = x;
    S.add(i + 1, i + 2, x);
  }
  for (long long i = 0; i < (long long)(q); i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      long double sum1 = 0, sum2 = 0;
      sum1 = S.getSum(l1, r1 + 1) / (r1 - l1 + 1);
      sum2 = S.getSum(l2, r2 + 1) / (r2 - l2 + 1);
      long double p1 = 1.0 / (r1 - l1 + 1);
      long double p2 = 1.0 / (r2 - l2 + 1);
      S.mulValue(l1, r1 + 1, 1 - p1);
      S.mulValue(l2, r2 + 1, 1 - p2);
      S.add(l1, r1 + 1, p1 * sum2);
      S.add(l2, r2 + 1, p2 * sum1);
    } else {
      int l, r;
      cin >> l >> r;
      cout << setprecision(10) << fixed << S.getSum(l, r + 1) << endl;
    }
  }
}