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 std::lower_bound;
using std::map;
using std::queue;
using std::sort;
using std::string;
using std::upper_bound;
using std::vector;
template <class T>
class SegmentTree {
 public:
  T dat, lazym, lazyp;
  int leftBorder, rightBorder, mid;
  SegmentTree *leftSon, *rightSon;
  SegmentTree() {
    leftBorder = rightBorder = -1;
    leftSon = rightSon = NULL;
  }
  void pushdown();
  void pushup();
  void Build(T *, int, int);
  void Modify(int, int, T, T);
  T Query(int, int);
  void Free();
};
template <class T>
void SegmentTree<T>::pushdown() {
  if ((lazym != 1 || lazyp != (T)0) && leftBorder != rightBorder) {
    leftSon->dat = leftSon->dat * lazym +
                   lazyp * (leftSon->rightBorder - leftSon->leftBorder + 1);
    rightSon->dat = rightSon->dat * lazym +
                    lazyp * (rightSon->rightBorder - rightSon->leftBorder + 1);
    leftSon->lazym *= lazym;
    leftSon->lazyp = leftSon->lazyp * lazym + lazyp;
    rightSon->lazym *= lazym;
    rightSon->lazyp = rightSon->lazyp * lazym + lazyp;
  }
  lazym = (T)1;
  lazyp = (T)0;
}
template <class T>
void SegmentTree<T>::pushup() {
  dat = leftSon->dat + rightSon->dat;
}
template <class T>
void SegmentTree<T>::Build(T *S, int l, int r) {
  if (l > r) {
    return;
  }
  lazym = (T)1;
  lazyp = (T)0;
  leftBorder = l;
  rightBorder = r;
  mid = (leftBorder + rightBorder) >> 1;
  if (l == r) {
    dat = S[l];
    return;
  }
  leftSon = new SegmentTree;
  leftSon->Build(S, l, mid);
  rightSon = new SegmentTree;
  rightSon->Build(S, mid + 1, r);
  pushup();
}
template <class T>
void SegmentTree<T>::Modify(int l, int r, T mult, T plus) {
  if (l > r || l < leftBorder || rightBorder < r) {
    return;
  }
  if (leftBorder == l && rightBorder == r) {
    dat = dat * mult + plus * (rightBorder - leftBorder + 1);
    lazym *= mult;
    lazyp = lazyp * mult + plus;
    return;
  }
  pushdown();
  if (r <= mid) {
    leftSon->Modify(l, r, mult, plus);
  } else if (mid < l) {
    rightSon->Modify(l, r, mult, plus);
  } else {
    leftSon->Modify(l, mid, mult, plus);
    rightSon->Modify(mid + 1, r, mult, plus);
  }
  pushup();
}
template <class T>
T SegmentTree<T>::Query(int l, int r) {
  if (l > r || l < leftBorder || rightBorder < r) {
    return dat;
  }
  pushdown();
  if (l == leftBorder && r == rightBorder) {
    return dat;
  }
  if (r <= mid) {
    return leftSon->Query(l, r);
  } else if (mid < l) {
    return rightSon->Query(l, r);
  } else {
    return leftSon->Query(l, mid) + rightSon->Query(mid + 1, r);
  }
}
template <class T>
void SegmentTree<T>::Free() {
  if (leftSon != NULL) {
    leftSon->Free();
  }
  if (rightSon != NULL) {
    rightSon->Free();
  }
  delete leftSon;
  delete rightSon;
}
int cmp(const void *x, const void *y) {
  int dx = *((int *)(x)), dy = *((int *)(y));
  return dx > dy ? 1 : -1;
}
SegmentTree<double> st;
double a[100005];
int main() {
  int n, q;
  scanf("%d", &n);
  scanf("%d", &q);
  for (int i = 0; i < n; i++) scanf("%lf", &a[i]);
  st.Build(a, 0, n - 1);
  int t, l1, l2, l, r1, r2, r;
  for (int i = 0; i < q; i++) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d", &l1);
      scanf("%d", &r1);
      scanf("%d", &l2);
      scanf("%d", &r2);
      l1--, r1--, l2--, r2--;
      double sum1 = st.Query(l1, r1), sum2 = st.Query(l2, r2);
      double ave1 = sum1 / (r1 - l1 + 1.0), ave2 = sum2 / (r2 - l2 + 1.0);
      st.Modify(l1, r1, (r1 - l1) / (r1 - l1 + 1.0), ave2 / (r1 - l1 + 1.0));
      st.Modify(l2, r2, (r2 - l2) / (r2 - l2 + 1.0), ave1 / (r2 - l2 + 1.0));
    } else {
      scanf("%d", &l);
      scanf("%d", &r);
      l--, r--;
      double ans = st.Query(l, r);
      printf("%.7lf\n", ans);
    }
  }
  return 0;
}