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;
struct node {
  double mean;
  double add;
  double mul;
  node() : mean(0), add(0), mul(1) {}
};
vector<node> t;
void push(int v) {
  t[v * 2].mean *= t[v].mul;
  t[v * 2].mul *= t[v].mul;
  t[v * 2].add *= t[v].mul;
  t[v * 2 + 1].mean *= t[v].mul;
  t[v * 2 + 1].mul *= t[v].mul;
  t[v * 2 + 1].add *= t[v].mul;
  t[v].mul = 1;
  t[v * 2].mean += t[v].add;
  t[v * 2].add += t[v].add;
  t[v * 2 + 1].mean += t[v].add;
  t[v * 2 + 1].add += t[v].add;
  t[v].add = 0;
}
void mul(int v, int tl, int tr, int l, int r, double val) {
  if (tl == l && tr == r) {
    t[v].mean *= val;
    t[v].mul *= val;
    t[v].add *= val;
    return;
  }
  push(v);
  int tm = (tl + tr) >> 1;
  if (l <= tm) {
    mul(v * 2, tl, tm, l, min(r, tm), val);
  }
  if (r > tm) {
    mul(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, val);
  }
  t[v].mean = (t[v * 2].mean * (tm - tl + 1) + t[v * 2 + 1].mean * (tr - tm)) /
              (tr - tl + 1);
}
void add(int v, int tl, int tr, int l, int r, double val) {
  if (tl == l && tr == r) {
    t[v].mean += val;
    t[v].add += val;
    return;
  }
  push(v);
  int tm = (tl + tr) >> 1;
  if (l <= tm) {
    add(v * 2, tl, tm, l, min(r, tm), val);
  }
  if (r > tm) {
    add(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, val);
  }
  t[v].mean = (t[v * 2].mean * (tm - tl + 1) + t[v * 2 + 1].mean * (tr - tm)) /
              (tr - tl + 1);
}
void build(int v, int tl, int tr, vector<double>& a) {
  if (tl == tr) {
    t[v].mean = a[tl];
    return;
  }
  int tm = (tl + tr) >> 1;
  build(v * 2, tl, tm, a);
  build(v * 2 + 1, tm + 1, tr, a);
  t[v].mean = (t[v * 2].mean * (tm - tl + 1) + t[v * 2 + 1].mean * (tr - tm)) /
              (tr - tl + 1);
}
double get(int v, int tl, int tr, int l, int r) {
  if (tl == l && tr == r) {
    return t[v].mean * (tr - tl + 1);
  }
  push(v);
  double ans = 0;
  int tm = (tl + tr) >> 1;
  if (l <= tm) {
    ans += get(v * 2, tl, tm, l, min(r, tm));
  }
  if (r > tm) {
    ans += get(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  vector<double> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  t.resize(4 * n);
  build(1, 0, n - 1, a);
  while (q--) {
    int type;
    cin >> type;
    if (type == 2) {
      int l, r;
      cin >> l >> r;
      --l, --r;
      double val = get(1, 0, n - 1, l, r);
      cout << fixed;
      cout.precision(20);
      cout << val << "\n";
    } else {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      --l1, --r1, --l2, --r2;
      int x = r1 - l1 + 1;
      int y = r2 - l2 + 1;
      double lf = get(1, 0, n - 1, l1, r1);
      double rg = get(1, 0, n - 1, l2, r2);
      mul(1, 0, n - 1, l1, r1, double(x - 1) / x);
      mul(1, 0, n - 1, l2, r2, double(y - 1) / y);
      add(1, 0, n - 1, l1, r1, rg / y / x);
      add(1, 0, n - 1, l2, r2, lf / y / x);
    }
  }
}