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 {
  bool d;
  double m = 1, a;
  double v;
};
node tree[1 << 18];
int n;
int in[100500];
double init(int pos, int l, int r) {
  if (l == r) {
    return tree[pos].v = in[l];
  }
  int mid = (l + r) / 2;
  return tree[pos].v = init(2 * pos, l, mid) + init(2 * pos + 1, mid + 1, r);
}
void push(int pos, int l, int r) {
  node& n = tree[pos];
  if (n.d) {
    if (l == r) {
      n.v = n.v * n.m + n.a;
    } else {
      node& c1 = tree[pos * 2];
      c1.a *= n.m;
      c1.m *= n.m;
      c1.a += n.a;
      node& c2 = tree[1 + pos * 2];
      c2.a *= n.m;
      c2.m *= n.m;
      c2.a += n.a;
      c1.d = c2.d = true;
      n.v = n.v * n.m + n.a * (r - l + 1);
    }
    n.d = false;
    n.m = 1;
    n.a = 0;
  }
}
void update(int pos, int l, int r, int ql, int qr, double M, double A) {
  push(pos, l, r);
  if (qr < l || ql > r) return;
  if (l == r) {
    tree[pos].v = tree[pos].v * M + A;
  } else {
    if (ql <= l && r <= qr) {
      node& n = tree[pos];
      n.a = n.a * M + A;
      n.m *= M;
      n.d = true;
    } else {
      int mid = (l + r) / 2;
      update(2 * pos, l, mid, ql, qr, M, A);
      update(2 * pos + 1, mid + 1, r, ql, qr, M, A);
      tree[pos].v = tree[2 * pos].v + tree[2 * pos + 1].v;
    }
  }
  push(pos, l, r);
}
double query(int pos, int l, int r, int ql, int qr) {
  push(pos, l, r);
  if (qr < l || ql > r) return 0;
  if (ql <= l && r <= qr) return tree[pos].v;
  int mid = (l + r) / 2;
  return query(2 * pos, l, mid, ql, qr) +
         query(2 * pos + 1, mid + 1, r, ql, qr);
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = (1); i <= (n); ++i) scanf("%d", &in[i]);
  init(1, 1, n);
  int tp;
  int a1, b1, a2, b2;
  while (q--) {
    scanf("%d", &tp);
    if (tp == 1) {
      scanf("%d %d", &a1, &b1);
      scanf("%d %d", &a2, &b2);
      double s1 = query(1, 1, n, a1, b1);
      double s2 = query(1, 1, n, a2, b2);
      int l1 = b1 - a1 + 1, l2 = b2 - a2 + 1;
      update(1, 1, n, a1, b1, (double)(l1 - 1) / l1, s2 / l2 / l1);
      update(1, 1, n, a2, b2, (double)(l2 - 1) / l2, s1 / l1 / l2);
    } else {
      scanf("%d %d", &a1, &b1);
      double res = query(1, 1, n, a1, b1);
      printf("%.9lf\n", res);
    }
  }
}