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 nmax = 100005;
struct nodee {
  double sum, s, p;
} arb[4 * nmax];
long double v[nmax];
int n, M, i, j, st, dr, st1, dr1, st2, dr2, tip;
long double ans, l1, l2, s1, s2, prod, ss;
string s;
int ch, num;
void update(int nod, int l, int r, int st, int dr, double P, double S);
void push(int nod, int l, int r) {
  int m = (l + r) / 2;
  if (l == r) {
    arb[nod].sum = (arb[nod].sum * arb[nod].p + arb[nod].s);
  } else {
    arb[2 * nod].p *= arb[nod].p;
    arb[2 * nod + 1].p *= arb[nod].p;
    arb[2 * nod].s = arb[2 * nod].s * arb[nod].p + arb[nod].s;
    arb[2 * nod + 1].s = arb[2 * nod + 1].s * arb[nod].p + arb[nod].s;
    double lst = m - l + 1, ldr = r - m;
    arb[nod].sum =
        (arb[2 * nod].sum * arb[2 * nod].p + lst * arb[2 * nod].s +
         arb[2 * nod + 1].sum * arb[2 * nod + 1].p + ldr * arb[2 * nod + 1].s);
  }
  arb[nod].p = 1;
  arb[nod].s = 0;
}
void build(int nod, int l, int r) {
  if (l == r) {
    arb[nod].sum = v[l];
    arb[nod].p = 1;
    arb[nod].s = 0;
    return;
  }
  int m = (l + r) / 2;
  build(2 * nod, l, m);
  build(2 * nod + 1, m + 1, r);
  arb[nod].p = 1;
  arb[nod].s = 0;
  arb[nod].sum = (arb[2 * nod].sum + arb[2 * nod + 1].sum);
  return;
}
void update(int nod, int l, int r, int st, int dr, double P, double S) {
  if (arb[nod].p != 1 || arb[nod].s != 0) push(nod, l, r);
  if (st <= l && r <= dr) {
    arb[nod].s = S;
    arb[nod].p = P;
    return;
  }
  int m = (l + r) / 2;
  if (st <= m) update(2 * nod, l, m, st, dr, P, S);
  if (m < dr) update(2 * nod + 1, m + 1, r, st, dr, P, S);
  double lst = m - l + 1, ldr = r - m;
  arb[nod].sum =
      (arb[2 * nod].sum * arb[2 * nod].p + lst * arb[2 * nod].s +
       arb[2 * nod + 1].sum * arb[2 * nod + 1].p + ldr * arb[2 * nod + 1].s);
}
void query(int nod, int l, int r) {
  if (st <= l && r <= dr) {
    double lll = (r - l + 1);
    ans += arb[nod].sum * arb[nod].p + lll * arb[nod].s;
    return;
  }
  if (arb[nod].p != 1 || arb[nod].s != 0) push(nod, l, r);
  int m = (l + r) / 2;
  if (st <= m) query(2 * nod, l, m);
  if (m < dr) query(2 * nod + 1, m + 1, r);
  double lst = m - l + 1, ldr = r - m;
  arb[nod].sum =
      (arb[2 * nod].sum * arb[2 * nod].p + lst * arb[2 * nod].s +
       arb[2 * nod + 1].sum * arb[2 * nod + 1].p + ldr * arb[2 * nod + 1].s);
}
int getnum() {
  num = 0;
  while (s[ch] >= '0' && s[ch] <= '9') {
    num = num * 10 + s[ch] - '0';
    ch++;
  }
  ch++;
  return num;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> M;
  getline(cin, s);
  getline(cin, s);
  num = ch = 0;
  for (i = 1; i <= n; i++) {
    v[i] = getnum();
  }
  build(1, 1, n);
  for (i = 1; i <= M; i++) {
    getline(cin, s);
    num = ch = 0;
    tip = getnum();
    if (tip == 1) {
      st1 = getnum();
      dr1 = getnum();
      st2 = getnum();
      dr2 = getnum();
      st = st1;
      dr = dr1;
      ans = 0;
      query(1, 1, n);
      l1 = dr1 - st1 + 1;
      s1 = ans;
      st = st2;
      dr = dr2;
      ans = 0;
      query(1, 1, n);
      l2 = dr2 - st2 + 1;
      s2 = ans;
      prod = (l1 - 1) / l1;
      ss = (s2 / l2) / l1;
      update(1, 1, n, st1, dr1, prod, ss);
      prod = (l2 - 1) / l2;
      ss = (s1 / l1) / l2;
      update(1, 1, n, st2, dr2, prod, ss);
    } else {
      st = getnum();
      dr = getnum();
      ans = 0;
      query(1, 1, n);
      cout << fixed << setprecision(5) << ans << '\n';
    }
  }
  return 0;
}