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;
double tree[4 * 300005];
pair<double, double> lazy[4 * 3000005];
int a[300005];
void build(int ind, int ss, int se) {
  if (ss > se) {
    return;
  }
  if (ss == se) {
    tree[ind] = a[ss];
    lazy[ind] = make_pair(-1, -1);
    return;
  }
  int mid = (ss + se) / 2;
  build(2 * ind + 1, ss, mid);
  build(2 * ind + 2, mid + 1, se);
  tree[ind] = tree[2 * ind + 1] + tree[2 * ind + 2];
  lazy[ind] = make_pair(-1, -1);
}
pair<double, double> addPair(pair<double, double> s1, pair<double, double> s2) {
  if (s1.first == -1) {
    return s2;
  }
  return make_pair(s1.first * s2.first, s1.second * s2.first + s2.second);
}
void update(int ind, int ss, int se, int qs, int qe, double mul, double add) {
  if (lazy[ind].first != -1) {
    tree[ind] = lazy[ind].first * tree[ind] + (se - ss + 1) * lazy[ind].second;
    if (ss != se) {
      lazy[2 * ind + 1] = addPair(lazy[2 * ind + 1], lazy[ind]);
      lazy[2 * ind + 2] = addPair(lazy[2 * ind + 2], lazy[ind]);
    }
    lazy[ind] = make_pair(-1, -1);
  }
  if (ss > se || qs > se || ss > qe) {
    return;
  }
  if (ss >= qs && se <= qe) {
    tree[ind] = mul * tree[ind] + (se - ss + 1) * add;
    if (ss != se) {
      lazy[2 * ind + 1] = addPair(lazy[2 * ind + 1], make_pair(mul, add));
      lazy[2 * ind + 2] = addPair(lazy[2 * ind + 2], make_pair(mul, add));
    }
    return;
  }
  int mid = (ss + se) / 2;
  update(2 * ind + 1, ss, mid, qs, qe, mul, add);
  update(2 * ind + 2, mid + 1, se, qs, qe, mul, add);
  tree[ind] = tree[2 * ind + 1] + tree[2 * ind + 2];
}
double getSum(int ind, int ss, int se, int qs, int qe) {
  if (lazy[ind].first != -1) {
    tree[ind] = lazy[ind].first * tree[ind] + (se - ss + 1) * lazy[ind].second;
    if (ss != se) {
      lazy[2 * ind + 1] = addPair(lazy[2 * ind + 1], lazy[ind]);
      lazy[2 * ind + 2] = addPair(lazy[2 * ind + 2], lazy[ind]);
    }
    lazy[ind] = make_pair(-1, -1);
  }
  if (ss > se || qs > se || ss > qe) {
    return 0.0;
  }
  if (ss >= qs && se <= qe) {
    return tree[ind];
  }
  int mid = (ss + se) / 2;
  return getSum(2 * ind + 1, ss, mid, qs, qe) +
         getSum(2 * ind + 2, mid + 1, se, qs, qe);
}
int main() {
  cout.setf(ios::fixed);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  build(0, 0, n - 1);
  for (int i = 0; i < 5 * n; i++) {
    lazy[i] = make_pair(-1, -1);
  }
  for (int i = 0; i < q; i++) {
    int fi;
    cin >> fi;
    if (fi == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      l1--;
      r1--;
      l2--;
      r2--;
      double sum1 = getSum(0, 0, n - 1, l1, r1);
      double sum2 = getSum(0, 0, n - 1, l2, r2);
      double avg1 = sum1 / (r1 - l1 + 1);
      double avg2 = sum2 / (r2 - l2 + 1);
      double size1 = (r1 - l1 + 1);
      double size2 = (r2 - l2 + 1);
      update(0, 0, n - 1, l1, r1, (1 - 1 / size1), avg2 / size1);
      update(0, 0, n - 1, l2, r2, (1 - 1 / size2), avg1 / size2);
    } else {
      int l1, r1;
      cin >> l1 >> r1;
      l1--;
      r1--;
      cout << std::setprecision(10) << getSum(0, 0, n - 1, l1, r1) << endl;
    }
  }
}