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 + 7, M = 1e6 + N / 2;
const long double eps = 0.2, eps2 = 0.0001;
const int INF = (int)1e9 + 7, MOD = INF;
const long long LINF = 1e16;
const double PI = acos(-1.0);
int n, a[N];
double mul[4 * N], add[4 * N];
double t[4 * N];
void build(int id, int l, int r) {
  mul[id] = 1;
  add[id] = 0;
  if (l == r) {
    t[id] = a[l];
    return;
  }
  int mid = l + r >> 1;
  int id2 = id + 1 << 1, id1 = id2 - 1;
  build(id1, l, mid), build(id2, mid + 1, r);
  t[id] = t[id1] + t[id2];
}
void push(int id) {
  int id2 = id + 1 << 1, id1 = id2 - 1;
  add[id2] = mul[id] * add[id2] + add[id];
  mul[id2] *= mul[id];
  add[id1] = mul[id] * add[id1] + add[id];
  mul[id1] *= mul[id];
  mul[id] = 1;
  add[id] = 0;
}
double get1(int id, int delta) { return t[id] * mul[id] + delta * add[id]; }
void update(int id, int l, int r, int L, int R, double mv, double av) {
  if (l > r) {
    return;
  }
  if (l == L && r == R) {
    add[id] *= mv;
    add[id] += av;
    mul[id] *= mv;
    return;
  }
  int mid = L + R >> 1;
  int id2 = id + 1 << 1, id1 = id2 - 1;
  push(id);
  update(id1, l, min(mid, r), L, mid, mv, av);
  update(id2, max(l, mid + 1), r, mid + 1, R, mv, av);
  t[id] = get1(id1, mid - L + 1) + get1(id2, R - mid);
}
double get(int id, int l, int r, int L, int R) {
  if (l > r) {
    return 0;
  }
  if (l == L && r == R) {
    return get1(id, R - L + 1);
  }
  int mid = L + R >> 1;
  int id2 = 1 + id << 1, id1 = id2 - 1;
  push(id);
  double ans = get(id1, l, min(mid, r), L, mid);
  ans += get(id2, max(l, mid + 1), r, mid + 1, R);
  t[id] = get1(id1, mid - L + 1) + get1(id2, R - mid);
  return ans;
}
int main() {
  int q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
  }
  build(0, 0, n - 1);
  for (int i = 0; i < q; i++) {
    int type, l1, r1;
    scanf("%d%d%d", &type, &l1, &r1);
    l1--, r1--;
    if (type == 2) {
      printf("%.7f\n", get(0, l1, r1, 0, n - 1));
      continue;
    }
    int l2, r2;
    scanf("%d%d", &l2, &r2);
    l2--, r2--;
    double av1 = get(0, l1, r1, 0, n - 1);
    av1 /= (r1 - l1 + 1);
    double av2 = get(0, l2, r2, 0, n - 1) / (r2 - l2 + 1);
    update(0, l2, r2, 0, n - 1, 1.0 * (r2 - l2) / (r2 - l2 + 1),
           av1 / (r2 - l2 + 1));
    update(0, l1, r1, 0, n - 1, 1.0 * (r1 - l1) / (r1 - l1 + 1),
           av2 / (r1 - l1 + 1));
  }
}