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 INF = 1e9;
const long double eps = 1e-9;
const int maxn = 1e5 + 10;
double seg[maxn << 2], Tag_Multi[maxn << 2], Tag_Plus[maxn << 2];
double ans = 0;
inline int read() {
  int x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') flag = -1, ch = getchar();
  while (isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * flag;
}
inline void Build(int h, int l, int r) {
  if (l == r) {
    seg[h] = 1.0 * read();
    return;
  }
  int mid = (l + r) >> 1;
  Build(h << 1, l, mid);
  Build(h << 1 | 1, mid + 1, r);
  seg[h] = seg[h << 1] + seg[h << 1 | 1];
}
inline void Pushdown_Multi(int h, double delta) {
  seg[h] *= delta;
  Tag_Multi[h] *= delta;
  Tag_Plus[h] *= delta;
}
inline void Pushdown_Plus(int h, int l, int r, double delta) {
  seg[h] += (r - l + 1) * delta;
  Tag_Plus[h] += delta;
}
inline void Pushdown(int h, int l, int r) {
  int mid = (l + r) >> 1;
  if (Tag_Multi[h] != 1) {
    Pushdown_Multi(h << 1, Tag_Multi[h]);
    Pushdown_Multi(h << 1 | 1, Tag_Multi[h]);
    Tag_Multi[h] = 1;
  }
  if (Tag_Plus[h] != 0) {
    Pushdown_Plus(h << 1, l, mid, Tag_Plus[h]);
    Pushdown_Plus(h << 1 | 1, mid + 1, r, Tag_Plus[h]);
    Tag_Plus[h] = 0;
  }
}
inline void Modify_Plus(int h, int L, int R, int l, int r, double delta) {
  if (l <= L && R <= r) {
    seg[h] += (R - L + 1) * delta;
    Tag_Plus[h] += delta;
    return;
  }
  int mid = (L + R) >> 1;
  Pushdown(h, L, R);
  if (l <= mid) Modify_Plus(h << 1, L, mid, l, r, delta);
  if (r >= mid + 1) Modify_Plus(h << 1 | 1, mid + 1, R, l, r, delta);
  seg[h] = seg[h << 1] + seg[h << 1 | 1];
}
inline void Modify_Multi(int h, int L, int R, int l, int r, double delta) {
  if (l <= L && R <= r) {
    seg[h] *= delta;
    Tag_Multi[h] *= delta;
    Tag_Plus[h] *= delta;
    return;
  }
  int mid = (L + R) >> 1;
  Pushdown(h, L, R);
  if (l <= mid) Modify_Multi(h << 1, L, mid, l, r, delta);
  if (r >= mid + 1) Modify_Multi(h << 1 | 1, mid + 1, R, l, r, delta);
  seg[h] = seg[h << 1] + seg[h << 1 | 1];
}
inline double Query(int h, int L, int R, int l, int r) {
  if (l <= L && R <= r) return seg[h];
  int mid = (L + R) >> 1;
  Pushdown(h, L, R);
  double curans = 0;
  if (l <= mid) curans += Query(h << 1, L, mid, l, r);
  if (r >= mid + 1) curans += Query(h << 1 | 1, mid + 1, R, l, r);
  return curans;
}
inline void Modify(int n, int l1, int r1, int l2, int r2) {
  double sum1 = Query(1, 1, n, l1, r1);
  double sum2 = Query(1, 1, n, l2, r2);
  sum1 = (sum1 * 1.0 / (r1 - l1 + 1)) * 1.0 / (r2 - l2 + 1);
  sum2 = (sum2 * 1.0 / (r1 - l1 + 1)) * 1.0 / (r2 - l2 + 1);
  Modify_Multi(1, 1, n, l1, r1, (r1 - l1) * 1.0 / (r1 - l1 + 1));
  Modify_Multi(1, 1, n, l2, r2, (r2 - l2) * 1.0 / (r2 - l2 + 1));
  Modify_Plus(1, 1, n, l1, r1, sum2);
  Modify_Plus(1, 1, n, l2, r2, sum1);
}
int main() {
  int i, j, n, q;
  n = read();
  q = read();
  Build(1, 1, n);
  for (i = 1; i <= (n << 2); i++) Tag_Multi[i] = 1.0;
  for (i = 1; i <= q; i++) {
    int opt = read();
    if (opt == 1) {
      int l1, r1, l2, r2;
      l1 = read();
      r1 = read();
      l2 = read();
      r2 = read();
      Modify(n, l1, r1, l2, r2);
    } else {
      int l, r;
      l = read();
      r = read();
      printf("%.8lf\n", Query(1, 1, n, l, r));
    }
  }
  return 0;
}