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;
const int maxN = 1e5 + 7;
struct node {
  int l, r;
  double sum, lazy, mul;
} tr[maxN << 2];
void init(int now, int l, int r) {
  tr[now].l = l;
  tr[now].r = r;
  tr[now].mul = 1.0;
  tr[now].lazy = 0;
  return;
}
void update(int now) {
  tr[now].sum = tr[now << 1].sum + tr[now << 1 | 1].sum;
  return;
}
void build(int now, int l, int r) {
  init(now, l, r);
  if (l == r) {
    double tp;
    scanf("%lf", &tp);
    tr[now].sum = tp;
    return;
  } else {
    int mid = (l + r) / 2;
    build(now << 1, l, mid);
    build(now << 1 | 1, mid + 1, r);
    update(now);
    return;
  }
}
void work(int now, double mul, double lazy) {
  tr[now].mul *= mul;
  tr[now].lazy = mul * tr[now].lazy + lazy;
  tr[now].sum = tr[now].sum * mul + lazy * (tr[now].r - tr[now].l + 1);
  return;
}
void pushdown(int now) {
  work(now << 1, tr[now].mul, tr[now].lazy);
  work(now << 1 | 1, tr[now].mul, tr[now].lazy);
  tr[now].mul = 1.0;
  tr[now].lazy = 0.0;
  return;
}
double query(int now, int L, int R) {
  if (tr[now].l == L && tr[now].r == R) {
    return tr[now].sum;
  } else {
    if (tr[now].mul != 1.0 || tr[now].lazy != 0.0) {
      pushdown(now);
    }
    int mid = (tr[now].l + tr[now].r) / 2;
    if (R <= mid)
      return query(now << 1, L, R);
    else if (L > mid)
      return query(now << 1 | 1, L, R);
    else
      return query(now << 1, L, mid) + query(now << 1 | 1, mid + 1, R);
  }
}
void modify(int now, int L, int R, double mul, double lazy) {
  if (tr[now].l == L && tr[now].r == R) {
    work(now, mul, lazy);
    return;
  } else {
    if (!(tr[now].mul == 1.0 && tr[now].lazy == 0.0)) pushdown(now);
    int mid = (tr[now].l + tr[now].r) / 2;
    if (R <= mid)
      modify(now << 1, L, R, mul, lazy);
    else if (L > mid)
      modify(now << 1 | 1, L, R, mul, lazy);
    else {
      modify(now << 1, L, mid, mul, lazy);
      modify(now << 1 | 1, mid + 1, R, mul, lazy);
    }
    update(now);
    return;
  }
}
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    build(1, 1, n);
    int opt, l1, r1, l2, r2;
    while (m--) {
      scanf("%d", &opt);
      scanf("%d%d", &l1, &r1);
      opt--;
      if (!opt) {
        scanf("%d%d", &l2, &r2);
        double t1 = (r1 - l1 + 1);
        double t2 = (r2 - l2 + 1);
        double s1 = query(1, l1, r1);
        double s2 = query(1, l2, r2);
        modify(1, l1, r1, (t1 - 1.0) / t1, s2 / (t1 * t2));
        modify(1, l2, r2, (t2 - 1.0) / t2, s1 / (t2 * t1));
      } else {
        printf("%.7f\n", query(1, l1, r1));
      }
    }
  }
  return 0;
}