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;
int n, q;
int a[100005];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
struct segment_tree {
  struct tree_node {
    double sum, ptag, mtag;
    int size;
  } tree[100005 * 4];
  void build(int p, int l, int r) {
    tree[p].size = r - l + 1, tree[p].mtag = 1;
    if (l == r) {
      tree[p].sum = a[l];
      return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid), build(p << 1 | 1, mid + 1, r);
    tree[p].sum = tree[p << 1].sum + tree[p << 1 | 1].sum;
  }
  void add_ptag(int p, double tag) {
    tree[p].ptag += tag, tree[p].sum += tree[p].size * tag;
  }
  void add_mtag(int p, double tag) {
    tree[p].mtag *= tag, tree[p].ptag *= tag, tree[p].sum *= tag;
  }
  void push_down(int p) {
    double &mt = tree[p].mtag, &pt = tree[p].ptag;
    if (mt) add_mtag(p << 1, mt), add_mtag(p << 1 | 1, mt), mt = 1;
    if (pt) add_ptag(p << 1, pt), add_ptag(p << 1 | 1, pt), pt = 0;
  }
  void change(int p, int l, int r, int x, int y, double mt, double pt) {
    if (x <= l && r <= y) {
      add_mtag(p, mt), add_ptag(p, pt);
      return;
    }
    int mid = (l + r) >> 1;
    push_down(p);
    if (x <= mid) change(p << 1, l, mid, x, y, mt, pt);
    if (y > mid) change(p << 1 | 1, mid + 1, r, x, y, mt, pt);
    tree[p].sum = tree[p << 1].sum + tree[p << 1 | 1].sum;
  }
  double query(int p, int l, int r, int x, int y) {
    if (x <= l && r <= y) return tree[p].sum;
    int mid = (l + r) >> 1;
    double ret = 0;
    push_down(p);
    if (x <= mid) ret += query(p << 1, l, mid, x, y);
    if (y > mid) ret += query(p << 1 | 1, mid + 1, r, x, y);
    return ret;
  }
  double query(int l, int r) { return query(1, 1, n, l, r); }
  void print(int p, int l, int r) {
    if (l == r) {
      printf("%.7f ", tree[p].sum);
      return;
    }
    push_down(p);
    int mid = (l + r) >> 1;
    print(p << 1, l, mid), print(p << 1 | 1, mid + 1, r);
  }
} T;
int main() {
  n = read(), q = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  T.build(1, 1, n);
  while (q--) {
    int type = read();
    if (type == 1) {
      int x1 = read(), y1 = read(), x2 = read(), y2 = read();
      int len1 = y1 - x1 + 1, len2 = y2 - x2 + 1;
      double sum1 = T.query(x1, y1), sum2 = T.query(x2, y2);
      T.change(1, 1, n, x1, y1, 1.0 * (len1 - 1) / len1, sum2 / len1 / len2);
      T.change(1, 1, n, x2, y2, 1.0 * (len2 - 1) / len2, sum1 / len1 / len2);
    } else {
      int l = read(), r = read();
      printf("%.7f\n", T.query(l, r));
    }
  }
  return 0;
}