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 gi() {
  char ch = '!';
  int res = 0, f = 0;
  while (ch < '0' || ch > '9') {
    ch = getchar();
    if (ch == '-') f = 1;
  }
  while (ch >= '0' && ch <= '9') res = res * 10 + ch - '0', ch = getchar();
  return f ? -res : res;
}
const int N = 1e5 + 100;
const double eps = 1e-10;
int n, Q, arr[N];
double sum[N << 2];
struct data {
  double a, b;
  friend data operator+(data A, data B) {
    return (data){A.a * B.a, A.b * B.a + B.b};
  }
} tag[N << 2];
void puttag(int k, data s, int l, int r) {
  sum[k] = sum[k] * s.a + s.b * (r - l + 1);
  tag[k] = tag[k] + s;
}
void pushdown(int k, int l, int r) {
  int mid = (l + r) >> 1;
  puttag((k << 1), tag[k], l, mid);
  puttag((k << 1 | 1), tag[k], mid + 1, r);
  tag[k] = (data){1, 0};
}
void init(int k = 1, int l = 1, int r = n) {
  tag[k] = (data){1, 0};
  if (l == r) return (void)(sum[k] = arr[l]);
  int mid = (l + r) >> 1;
  init((k << 1), l, mid), init((k << 1 | 1), mid + 1, r);
  sum[k] = sum[(k << 1)] + sum[(k << 1 | 1)];
}
void Modify(int k, int ql, int qr, data s, int l = 1, int r = n) {
  if (ql <= l && r <= qr) {
    puttag(k, s, l, r);
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(k, l, r);
  if (ql <= mid) Modify((k << 1), ql, qr, s, l, mid);
  if (qr > mid) Modify((k << 1 | 1), ql, qr, s, mid + 1, r);
  sum[k] = sum[(k << 1)] + sum[(k << 1 | 1)];
}
double Query(int k, int ql, int qr, int l = 1, int r = n) {
  if (ql <= l && r <= qr) {
    return sum[k];
  }
  int mid = (l + r) >> 1;
  double res = 0;
  pushdown(k, l, r);
  if (ql <= mid) res += Query((k << 1), ql, qr, l, mid);
  if (qr > mid) res += Query((k << 1 | 1), ql, qr, mid + 1, r);
  return res;
}
int main() {
  n = gi(), Q = gi();
  for (int i = 1; i <= n; ++i) arr[i] = gi();
  init();
  while (Q--) {
    int op = gi();
    if (op == 1) {
      int l1 = gi(), r1 = gi(), l2 = gi(), r2 = gi();
      double sum1 = Query(1, l1, r1), sum2 = Query(1, l2, r2);
      sum1 /= (double)(r1 - l1 + 1) * (r2 - l2 + 1);
      sum2 /= (double)(r1 - l1 + 1) * (r2 - l2 + 1);
      Modify(1, l1, r1, (data){(r1 - l1) / (double)(r1 - l1 + 1), 0});
      Modify(1, l1, r1, (data){1, sum2});
      Modify(1, l2, r2, (data){(r2 - l2) / (double)(r2 - l2 + 1), sum1});
    }
    if (op == 2) {
      int l = gi(), r = gi();
      printf("%.11lf\n", Query(1, l, r));
    }
  }
  return 0;
}