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, m;
int a[100005];
struct Tree {
  int left, right;
  double num;
  double chen;
  double jia;
} tree[500005];
void build(int left, int right, int i) {
  tree[i].left = left;
  tree[i].right = right;
  tree[i].chen = 1;
  tree[i].jia = 0;
  if (left == right) {
    tree[i].num = (double)a[left];
    tree[i].chen = 1;
    tree[i].jia = 0;
    return;
  }
  int mid = (left + right) / 2;
  build(left, mid, i * 2);
  build(mid + 1, right, i * 2 + 1);
  tree[i].num = tree[i * 2].num + tree[i * 2 + 1].num;
}
void Init() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build(1, n, 1);
}
void pushdown(int i, int s) {
  if (tree[i].chen != 1 || tree[i].jia != 0) {
    tree[i * 2].chen = tree[i * 2].chen * tree[i].chen;
    tree[i * 2 + 1].chen = tree[i * 2 + 1].chen * tree[i].chen;
    tree[i * 2].jia = tree[i * 2].jia * tree[i].chen + tree[i].jia;
    tree[i * 2 + 1].jia = tree[i * 2 + 1].jia * tree[i].chen + tree[i].jia;
    tree[i * 2].num =
        tree[i * 2].num * tree[i].chen + tree[i].jia * (double)(s - s / 2);
    tree[i * 2 + 1].num =
        tree[i * 2 + 1].num * tree[i].chen + tree[i].jia * (double)(s / 2);
    tree[i].chen = 1;
    tree[i].jia = 0;
  }
}
void pushup(int i) { tree[i].num = tree[i * 2].num + tree[i * 2 + 1].num; }
void Add(int left, int right, double c, double j, int i) {
  if (left == tree[i].left && right == tree[i].right) {
    tree[i].chen = tree[i].chen * c;
    tree[i].jia = tree[i].jia * c + j;
    tree[i].num = tree[i].num * c + j * (double)(right - left + 1);
    return;
  }
  if (tree[i].left == tree[i].right) return;
  pushdown(i, tree[i].right - tree[i].left + 1);
  int mid = (tree[i].left + tree[i].right) / 2;
  if (mid >= right) Add(left, right, c, j, i * 2);
  if (mid + 1 <= left) Add(left, right, c, j, i * 2 + 1);
  if (mid >= left && mid + 1 <= right) {
    Add(left, mid, c, j, i * 2);
    Add(mid + 1, right, c, j, i * 2 + 1);
  }
  pushup(i);
}
double Find(int left, int right, int i) {
  if (tree[i].left == left && tree[i].right == right) return tree[i].num;
  pushdown(i, tree[i].right - tree[i].left + 1);
  int mid = (tree[i].left + tree[i].right) / 2;
  if (mid >= right) return Find(left, right, i * 2);
  if (mid + 1 <= left) return Find(left, right, i * 2 + 1);
  if (mid >= left && mid + 1 <= right)
    return Find(left, mid, i * 2) + Find(mid + 1, right, i * 2 + 1);
}
void work() {
  int p;
  for (int i = 1; i <= m; i++) {
    scanf("%d", &p);
    if (p == 1) {
      int l1, r1, l2, r2;
      scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
      double sum1, sum2;
      sum1 = Find(l1, r1, 1);
      sum2 = Find(l2, r2, 1);
      Add(l1, r1, (double)(r1 - l1) / (double)(r1 - l1 + 1),
          sum2 / (double)(r1 - l1 + 1) / (double)(r2 - l2 + 1), 1);
      Add(l2, r2, (double)(r2 - l2) / (double)(r2 - l2 + 1),
          sum1 / (double)(r1 - l1 + 1) / (double)(r2 - l2 + 1), 1);
    }
    if (p == 2) {
      int l, r;
      scanf("%d %d", &l, &r);
      printf("%.7lf\n", Find(l, r, 1));
    }
  }
}
int main() {
  Init();
  work();
  return 0;
}