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;
const int MAXN = 4e5 + 10;
const int ADD = 69;
const int MUL = 420;
int n, q;
long double a[MAXN];
long double st[MAXN << 2];
long double mul[MAXN << 2], add[MAXN << 2];
void plant(int v, int b, int e) {
  mul[v] = 1;
  add[v] = 0;
  if (e - b == 1) {
    st[v] = a[b];
    return;
  }
  int m = (b + e) >> 1;
  plant(v << 1, b, m);
  plant(v << 1 ^ 1, m, e);
  st[v] = st[v << 1] + st[v << 1 ^ 1];
}
void nut(int v, int b, int e, int t, long double val) {
  if (t == ADD) {
    st[v] += (e - b) * val;
    add[v] += val;
  } else {
    st[v] *= val;
    add[v] *= val;
    mul[v] *= val;
  }
}
void weed(int v, int b, int e) {
  int m = (b + e) >> 1;
  nut(v << 1, b, m, MUL, mul[v]);
  nut(v << 1, b, m, ADD, add[v]);
  nut(v << 1 ^ 1, m, e, MUL, mul[v]);
  nut(v << 1 ^ 1, m, e, ADD, add[v]);
  add[v] = 0;
  mul[v] = 1;
}
void water(int v, int b, int e, int l, int r, int t, long double val) {
  if (l >= r) return;
  if (l == b && r == e) {
    nut(v, b, e, t, val);
    return;
  }
  weed(v, b, e);
  int m = (b + e) >> 1;
  water(v << 1, b, m, l, min(r, m), t, val);
  water(v << 1 ^ 1, m, e, max(m, l), r, t, val);
  st[v] = st[v << 1] + st[v << 1 ^ 1];
}
long double smoke(int v, int b, int e, int l, int r) {
  if (l >= r) return 0;
  if (l == b && e == r) {
    return st[v];
  }
  weed(v, b, e);
  int m = (b + e) >> 1;
  return smoke(v << 1, b, m, l, min(r, m)) +
         smoke(v << 1 ^ 1, m, e, max(m, l), r);
}
void solve() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  plant(1, 0, n);
  while (q--) {
    int qt;
    cin >> qt;
    if (qt == 1) {
      int l[2], r[2];
      for (int i = 0; i < 2; i++) cin >> l[i] >> r[i], --l[i];
      int len[2];
      for (int i = 0; i < 2; i++) len[i] = r[i] - l[i];
      long double s[2];
      for (int i = 0; i < 2; i++) s[i] = smoke(1, 0, n, l[i], r[i]);
      for (int i = 0; i < 2; i++)
        water(1, 0, n, l[i], r[i], MUL, (long double)(len[i] - 1) / len[i]);
      for (int i = 0; i < 2; i++)
        water(1, 0, n, l[i], r[i], ADD, s[i ^ 1] / len[0] / len[1]);
    } else {
      int l, r;
      cin >> l >> r;
      --l;
      long double s = smoke(1, 0, n, l, r);
      cout << s << "\n";
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(6);
  int te = 1;
  for (int w = 1; w <= te; w++) {
    solve();
  }
  return 0;
}