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;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long poww(long long a, long long b) {
  if (b == 0) return 1;
  long long tmp = poww(a, b / 2);
  return (b & 1 ? a * tmp * tmp : tmp * tmp);
}
string itos(long long i) {
  string s = "";
  while (i) {
    s += char(i % 10 + '0');
    i /= 10;
  }
  reverse(s.begin(), s.end());
  return s;
}
long long stoi(string &s) {
  long long tot = 0;
  for (int i = (int)s.length() - 1, j = 1; i >= 0; i--, j *= 10) {
    tot += j * (s[i] + '0');
  }
  return tot;
}
int months[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
using namespace std;
long long mod = 1e9 + 7;
void tt() { freopen("test.txt", "r", stdin); }
const int MAX = 1e6 + 10;
double a[MAX];
double st[4 * MAX], lazy[4 * MAX], add[4 * MAX], mul[4 * MAX];
void build(int node, int se, int en) {
  if (se == en) {
    st[node] = a[se];
    mul[node] = 1;
    return;
  }
  int mi = (se + en) / 2;
  build(node * 2, se, mi);
  build(node * 2 + 1, mi + 1, en);
  st[node] = st[node * 2] + st[node * 2 + 1];
  mul[node] = 1;
}
void push(int node, int se, int en) {
  int mi = (se + en) / 2;
  st[node * 2] = mul[node] * st[node * 2] + (double)(mi - se + 1) * add[node];
  st[node * 2 + 1] =
      mul[node] * st[node * 2 + 1] + (double)(en - mi) * add[node];
  st[node] = st[node * 2] + st[node * 2 + 1];
  mul[node * 2] *= mul[node];
  mul[node * 2 + 1] *= mul[node];
  add[node * 2] *= mul[node];
  add[node * 2 + 1] *= mul[node];
  add[node * 2] += add[node];
  add[node * 2 + 1] += add[node];
  add[node] = 0;
  mul[node] = 1;
}
double get(int node, int se, int en, int l, int r) {
  if (l > en || r < se) {
    return 0;
  }
  if (l <= se && en <= r) {
    return st[node];
  }
  push(node, se, en);
  int mi = (se + en) / 2;
  double p1 = get(node * 2, se, mi, l, r);
  double p2 = get(node * 2 + 1, mi + 1, en, l, r);
  return (p1 + p2);
}
void update(int node, int se, int en, int l, int r, double tomul,
            double toadd) {
  if (l > en || r < se) {
    return;
  }
  if (l <= se && en <= r) {
    st[node] *= tomul;
    st[node] += double(en - se + 1) * toadd;
    mul[node] *= tomul;
    add[node] *= tomul;
    add[node] += toadd;
    return;
  }
  int mi = (se + en) / 2;
  push(node, se, en);
  update(node * 2, se, mi, l, r, tomul, toadd);
  update(node * 2 + 1, mi + 1, en, l, r, tomul, toadd);
  st[node] = st[node * 2] + st[node * 2 + 1];
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  while (q--) {
    int type, l, r;
    cin >> type >> l >> r;
    if (type == 1) {
      int le, ri;
      cin >> le >> ri;
      double avg_val1 = get(1, 1, n, l, r) / double(r - l + 1);
      double avg_val2 = get(1, 1, n, le, ri) / double(ri - le + 1);
      update(1, 1, n, l, r, double(r - l) / double(r - l + 1),
             avg_val2 / double(r - l + 1));
      update(1, 1, n, le, ri, double(ri - le) / double(ri - le + 1),
             avg_val1 / double(ri - le + 1));
    } else
      cout << setprecision(10) << get(1, 1, n, l, r) << '\n';
  }
  return 0;
}