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>
inline int read() {
  register int x = 0, f = 1;
  register char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
struct SegmentTree {
  int n;
  double mul[400005], add[400005], sum[400005];
  inline void build(int p, int l, int r, int *a) {
    mul[p] = 1;
    add[p] = 0;
    if (l == r) {
      sum[p] = a[l];
      return;
    }
    int mid = l + r >> 1;
    build(p << 1, l, mid, a);
    build(p << 1 | 1, mid + 1, r, a);
    sum[p] = sum[p << 1] + sum[p << 1 | 1];
  }
  inline void init(int n, int *a) {
    this->n = n;
    build(1, 1, n, a);
  }
  inline void spread(int p, int l, int r) {
    int mid = l + r >> 1;
    if (mul[p] != 1) {
      mul[p << 1] *= mul[p];
      mul[p << 1 | 1] *= mul[p];
      add[p << 1] *= mul[p];
      add[p << 1 | 1] *= mul[p];
      sum[p << 1] *= mul[p];
      sum[p << 1 | 1] *= mul[p];
      mul[p] = 1;
    }
    if (add[p]) {
      add[p << 1] += add[p];
      add[p << 1 | 1] += add[p];
      sum[p << 1] += add[p] * (mid - l + 1);
      sum[p << 1 | 1] += add[p] * (r - mid);
      add[p] = 0;
    }
  }
  inline void _modify_mul(int p, int l, int r, int L, int R, double val) {
    if (L <= l && r <= R) {
      add[p] *= val;
      mul[p] *= val;
      sum[p] *= val;
      return;
    }
    int mid = l + r >> 1;
    spread(p, l, r);
    if (L <= mid) _modify_mul(p << 1, l, mid, L, R, val);
    if (R > mid) _modify_mul(p << 1 | 1, mid + 1, r, L, R, val);
    sum[p] = sum[p << 1] + sum[p << 1 | 1];
  }
  inline void _modify_add(int p, int l, int r, int L, int R, double val) {
    if (L <= l && r <= R) {
      add[p] += val;
      sum[p] += (r - l + 1) * val;
      return;
    }
    int mid = l + r >> 1;
    spread(p, l, r);
    if (L <= mid) _modify_add(p << 1, l, mid, L, R, val);
    if (R > mid) _modify_add(p << 1 | 1, mid + 1, r, L, R, val);
    sum[p] = sum[p << 1] + sum[p << 1 | 1];
  }
  inline double _ask(int p, int l, int r, int L, int R) {
    if (L <= l && r <= R) return sum[p];
    int mid = l + r >> 1;
    spread(p, l, r);
    if (R <= mid) return _ask(p << 1, l, mid, L, R);
    if (L > mid) return _ask(p << 1 | 1, mid + 1, r, L, R);
    return _ask(p << 1, l, mid, L, mid) +
           _ask(p << 1 | 1, mid + 1, r, mid + 1, R);
  }
  inline void modify(int opt, int L, int R, double val) {
    if (opt == 0)
      _modify_mul(1, 1, n, L, R, val);
    else
      _modify_add(1, 1, n, L, R, val);
  }
  inline double ask(int L, int R) { return _ask(1, 1, n, L, R); }
} SGT;
int a[100005];
int main() {
  int n = read(), Q = read();
  for (register int i = 1; i <= n; ++i) a[i] = read();
  SGT.init(n, a);
  while (Q--) {
    int op = read();
    if (op == 1) {
      int l1 = read(), r1 = read(), l2 = read(), r2 = read();
      int len1 = r1 - l1 + 1, len2 = r2 - l2 + 1;
      double sum1 = SGT.ask(l1, r1), sum2 = SGT.ask(l2, r2);
      SGT.modify(0, l1, r1, (len1 - 1) / (double)len1);
      SGT.modify(1, l1, r1, sum2 / len1 / len2);
      SGT.modify(0, l2, r2, (len2 - 1) / (double)len2);
      SGT.modify(1, l2, r2, sum1 / len1 / len2);
    } else {
      int l = read(), r = read();
      printf("%.5lf\n", SGT.ask(l, r));
    }
  }
  return 0;
}