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;
template <typename T>
inline bool cmin(T& a, const T& b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool cmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
int read() {
  int x = 0, f = 1;
  char ch;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  return x * f;
}
const int MaxN = 101234;
int n, q;
double seg[MaxN << 2], tag[MaxN << 2], add[MaxN << 2], mul[MaxN << 2];
void push_down(int p, int l, int r) {
  int mid = l + r >> 1;
  if (l == r) return;
  mul[p << 1] *= mul[p];
  mul[p << 1 | 1] *= mul[p];
  add[p << 1] = add[p << 1] * mul[p] + add[p];
  add[p << 1 | 1] = add[p << 1 | 1] * mul[p] + add[p];
  seg[p << 1] = seg[p << 1] * mul[p] + add[p] * (mid - l + 1);
  seg[p << 1 | 1] = seg[p << 1 | 1] * mul[p] + add[p] * (r - mid);
  mul[p] = 1;
  add[p] = 0;
}
void modify(int p, int l, int r, int a, int b, double x, double y) {
  if (l > b || r < a) return;
  if (a <= l && r <= b) {
    mul[p] *= x;
    add[p] = add[p] * x + y;
    seg[p] = seg[p] * x + (r - l + 1) * y;
    return;
  }
  push_down(p, l, r);
  modify(p << 1, l, l + r >> 1, a, b, x, y);
  modify(p << 1 | 1, (l + r >> 1) + 1, r, a, b, x, y);
  seg[p] = seg[p << 1] + seg[p << 1 | 1];
}
double query(int p, int l, int r, int a, int b) {
  if (l > b || r < a) return 0;
  if (a <= l && r <= b) return seg[p];
  push_down(p, l, r);
  return query(p << 1, l, l + r >> 1, a, b) +
         query(p << 1 | 1, (l + r >> 1) + 1, r, a, b);
}
void build(int p, int l, int r) {
  mul[p] = 1;
  if (l == r)
    seg[p] = read();
  else {
    build(p << 1, l, l + r >> 1);
    build(p << 1 | 1, (l + r >> 1) + 1, r);
    seg[p] = seg[p << 1] + seg[p << 1 | 1];
  }
}
int main() {
  int i;
  n = read();
  q = read();
  build(1, 1, n);
  while (q--) {
    int op = read();
    if (op == 1) {
      int l1 = read(), r1 = read(), l2 = read(), r2 = read();
      double x = query(1, 1, n, l1, r1) / (r1 - l1 + 1);
      double y = query(1, 1, n, l2, r2) / (r2 - l2 + 1);
      modify(1, 1, n, l1, r1, (double)(r1 - l1) / (r1 - l1 + 1),
             y / (r1 - l1 + 1));
      modify(1, 1, n, l2, r2, (double)(r2 - l2) / (r2 - l2 + 1),
             x / (r2 - l2 + 1));
    } else {
      int l = read(), r = read();
      printf("%.6lf\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}