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;
struct Node {
  double a, b, sum;
  int len;
} node[100010 << 2];
void init(int i, int l, int r) {
  node[i].a = 1;
  node[i].b = 0;
  node[i].len = r - l + 1;
  if (l == r) {
    scanf("%lf", &node[i].sum);
    return;
  }
  int mid = (l + r) >> 1;
  init(i << 1, l, mid);
  init((i << 1) | 1, mid + 1, r);
  node[i].sum = node[i << 1].sum + node[(i << 1) | 1].sum;
}
void pushdown(int i) {
  if (node[i].a != 1 || node[i].b != 0) {
    double a = node[i].a, b = node[i].b;
    node[i << 1].a *= a;
    node[i << 1].b = node[i << 1].b * a + b;
    node[i << 1].sum = node[i << 1].sum * a + b * node[i << 1].len;
    node[(i << 1) | 1].a *= a;
    node[(i << 1) | 1].b = node[(i << 1) | 1].b * a + b;
    node[(i << 1) | 1].sum =
        node[(i << 1) | 1].sum * a + b * node[(i << 1) | 1].len;
    node[i].a = 1;
    node[i].b = 0;
  }
}
void update(int i, int l, int r, int x, int y, double a, double b) {
  if (x == l && y == r) {
    node[i].a *= a;
    node[i].b = node[i].b * a + b;
    node[i].sum = node[i].sum * a + b * node[i].len;
    return;
  }
  pushdown(i);
  int mid = (l + r) >> 1;
  if (y <= mid)
    update(i << 1, l, mid, x, y, a, b);
  else if (x > mid)
    update((i << 1) | 1, mid + 1, r, x, y, a, b);
  else {
    update(i << 1, l, mid, x, mid, a, b);
    update((i << 1) | 1, mid + 1, r, mid + 1, y, a, b);
  }
  node[i].sum = node[i << 1].sum + node[(i << 1) | 1].sum;
}
double query(int i, int l, int r, int x, int y) {
  if (x == l && y == r) return node[i].sum;
  pushdown(i);
  int mid = (l + r) >> 1;
  if (y <= mid) return query(i << 1, l, mid, x, y);
  if (x > mid) return query((i << 1) | 1, mid + 1, r, x, y);
  return query(i << 1, l, mid, x, mid) +
         query((i << 1) | 1, mid + 1, r, mid + 1, y);
}
int main() {
  int n, q, type, l1, l2, r1, r2, l, r;
  scanf("%d%d", &n, &q);
  init(1, 1, n);
  while (q--) {
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double sum1 = query(1, 1, n, l1, r1);
      double sum2 = query(1, 1, n, l2, r2);
      update(1, 1, n, l1, r1, 1. * (r1 - l1) / (r1 - l1 + 1),
             1. / (r1 - l1 + 1) * sum2 / (r2 - l2 + 1));
      update(1, 1, n, l2, r2, 1. * (r2 - l2) / (r2 - l2 + 1),
             1. / (r2 - l2 + 1) * sum1 / (r1 - l1 + 1));
    } else {
      scanf("%d%d", &l, &r);
      printf("%.9f\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}