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 = 100050;
const double eps = 1e-6;
double a[maxn], sum[maxn << 2];
double addv[maxn << 2], multv[maxn << 2];
int dcmp(double x) {
  if (fabs(x) < eps) return 0;
  return (x < 0 ? -1 : 1);
}
void push_up(int p) { sum[p] = sum[p << 1] + sum[p << 1 | 1]; }
void push_down(int p, int l, int r) {
  if (dcmp(multv[p]) != 1 || dcmp(addv[p]) != 0) {
    multv[p << 1] *= multv[p];
    multv[p << 1 | 1] *= multv[p];
    addv[p << 1] = multv[p] * addv[p << 1] + addv[p];
    addv[p << 1 | 1] = multv[p] * addv[p << 1 | 1] + addv[p];
    int mid = (l + r) >> 1;
    sum[p << 1] = multv[p] * sum[p << 1] + addv[p] * (mid - l + 1);
    sum[p << 1 | 1] = multv[p] * sum[p << 1 | 1] + addv[p] * (r - mid);
    multv[p] = 1;
    addv[p] = 0;
  }
}
void build(int p, int l, int r) {
  multv[p] = 1;
  addv[p] = 0;
  if (l == r) {
    sum[p] = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  push_up(p);
}
void modify(int p, int l, int r, int L, int R, double k, double b) {
  if (L <= l && r <= R) {
    sum[p] = k * sum[p] + b * (r - l + 1);
    multv[p] *= k;
    addv[p] = k * addv[p] + b;
    return;
  }
  push_down(p, l, r);
  int mid = (l + r) >> 1;
  if (L <= mid) modify(p << 1, l, mid, L, R, k, b);
  if (R > mid) modify(p << 1 | 1, mid + 1, r, L, R, k, b);
  push_up(p);
}
double query(int p, int l, int r, int L, int R) {
  if (L <= l && r <= R) return sum[p];
  push_down(p, l, r);
  double ret = 0;
  int mid = (l + r) >> 1;
  if (L <= mid) ret += query(p << 1, l, mid, L, R);
  if (R > mid) ret += query(p << 1 | 1, mid + 1, r, L, R);
  return ret;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%lf", &a[i]);
  build(1, 1, n);
  while (q--) {
    int op, l1, r1, l2, r2;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double size1 = r1 - l1 + 1, size2 = r2 - l2 + 1;
      double t1 = query(1, 1, n, l1, r1), t2 = query(1, 1, n, l2, r2);
      modify(1, 1, n, l1, r1, (size1 - 1) / size1, t2 / (size2 * size1));
      modify(1, 1, n, l2, r2, (size2 - 1) / size2, t1 / (size1 * size2));
    } else {
      scanf("%d%d", &l1, &l2);
      double ans = query(1, 1, n, l1, l2);
      printf("%f\n", ans);
    }
  }
  return 0;
}