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 N = 1e5 + 5;
double seg[N << 2], mul[N << 2], add[N << 2];
int n, q;
void build(int l, int r, int rt) {
  seg[rt] = add[rt] = 0;
  mul[rt] = 1;
  if (l == r) {
    scanf("%lf", &seg[rt]);
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  seg[rt] = seg[rt << 1] + seg[rt << 1 | 1];
}
void push_down(int rt, int l, int r) {
  double _mul = mul[rt], _add = add[rt];
  int m = (l + r) >> 1;
  add[rt] = 0;
  mul[rt] = 1;
  add[rt << 1] = add[rt << 1] * _mul + _add;
  add[rt << 1 | 1] = add[rt << 1 | 1] * _mul + _add;
  mul[rt << 1] *= _mul;
  mul[rt << 1 | 1] *= _mul;
  seg[rt << 1] *= _mul;
  seg[rt << 1] += (m - l + 1) * _add;
  seg[rt << 1 | 1] *= _mul;
  seg[rt << 1 | 1] += (r - m) * _add;
}
void update(int L, int R, int l, int r, int rt, double _mul, double _add) {
  if (l >= L && r <= R) {
    seg[rt] *= _mul;
    seg[rt] += (r - l + 1) * _add;
    add[rt] = add[rt] * _mul + _add;
    mul[rt] *= _mul;
    return;
  }
  push_down(rt, l, r);
  int m = (l + r) >> 1;
  if (L <= m) update(L, R, l, m, rt << 1, _mul, _add);
  if (R > m) update(L, R, m + 1, r, rt << 1 | 1, _mul, _add);
  seg[rt] = seg[rt << 1] + seg[rt << 1 | 1];
}
double query(int L, int R, int l, int r, int rt) {
  if (l >= L && r <= R) return seg[rt];
  push_down(rt, l, r);
  int m = (l + r) >> 1;
  double ans = 0;
  if (L <= m) ans += query(L, R, l, m, rt << 1);
  if (R > m) ans += query(L, R, m + 1, r, rt << 1 | 1);
  return ans;
}
int main() {
  scanf("%d%d", &n, &q);
  build(1, n, 1);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l1, l2, r1, r2;
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double res1 = query(l1, r1, 1, n, 1) / (r1 - l1 + 1);
      double res2 = query(l2, r2, 1, n, 1) / (r2 - l2 + 1);
      update(l1, r1, 1, n, 1, 1.0 * (r1 - l1) / (r1 - l1 + 1),
             res2 / (r1 - l1 + 1));
      update(l2, r2, 1, n, 1, 1.0 * (r2 - l2) / (r2 - l2 + 1),
             res1 / (r2 - l2 + 1));
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      double ans = query(l, r, 1, n, 1);
      printf("%f\n", ans);
    }
  }
  return 0;
}