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 = 1e5 + 10;
const long long mod = 1e9 + 7;
const double eps = 1e-7;
double tree[maxn << 2];
double mult[maxn << 2], add[maxn << 2];
void pushup(int rt) {
  tree[rt] = tree[rt << 1] + tree[(rt << 1) | 1];
  mult[rt] = 1;
  add[rt] = 0;
  return;
}
void build(int rt, int l, int r) {
  if (l == r) {
    scanf("%lf", &tree[rt]);
    add[rt] = 0;
    mult[rt] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build((rt << 1) | 1, mid + 1, r);
  pushup(rt);
}
void pushdown(int rt, int l, int r) {
  tree[rt << 1] *= mult[rt];
  tree[(rt << 1) | 1] *= mult[rt];
  mult[rt << 1] *= mult[rt];
  mult[(rt << 1) | 1] *= mult[rt];
  add[rt << 1] *= mult[rt];
  add[(rt << 1) | 1] *= mult[rt];
  mult[rt] = 1;
  int mid = (l + r) >> 1;
  tree[rt << 1] += add[rt] * (mid - l + 1);
  tree[(rt << 1) | 1] += add[rt] * (r - mid);
  add[rt << 1] += add[rt];
  add[(rt << 1) | 1] += add[rt];
  add[rt] = 0;
  tree[rt] = tree[rt << 1] + tree[(rt << 1) | 1];
  return;
}
void update(int rt, int l, int r, int L, int R, double mulval, double addval) {
  if (l >= L && r <= R) {
    tree[rt] *= mulval;
    tree[rt] += addval * (r - l + 1);
    mult[rt] *= mulval;
    add[rt] *= mulval;
    add[rt] += addval;
    return;
  }
  pushdown(rt, l, r);
  int mid = (l + r) >> 1;
  if (L <= mid) update(rt << 1, l, mid, L, R, mulval, addval);
  if (R > mid) update((rt << 1) | 1, mid + 1, r, L, R, mulval, addval);
  tree[rt] = tree[rt << 1] + tree[(rt << 1) | 1];
  return;
}
double query(int rt, int l, int r, int L, int R) {
  if (l >= L && r <= R) return tree[rt];
  pushdown(rt, l, r);
  double ans = 0;
  int mid = (l + r) >> 1;
  if (L <= mid) ans += query(rt << 1, l, mid, L, R);
  if (R > mid) ans += query((rt << 1) | 1, mid + 1, r, L, R);
  return ans;
}
int ord;
int n, q;
int l1, r1, l2, r2;
int main() {
  cin >> n >> q;
  build(1, 1, n);
  while (q--) {
    cin >> ord;
    if (ord == 1) {
      cin >> l1 >> r1 >> l2 >> r2;
      double val1 = query(1, 1, n, l1, r1) / (r1 - l1 + 1);
      double val2 = query(1, 1, n, l2, r2) / (r2 - l2 + 1);
      update(1, 1, n, l1, r1, (1.0 * r1 - l1) / (r1 - l1 + 1),
             val2 / (r1 - l1 + 1));
      update(1, 1, n, l2, r2, (1.0 * r2 - l2) / (r2 - l2 + 1),
             val1 / (r2 - l2 + 1));
    } else {
      int l, r;
      cin >> l >> r;
      printf("%.10f\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}