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 N = 100010 << 4;
double s;
int L[N], R[N];
double sum[N], l1[N], l2[N], a[N];
void up(int t) { sum[t] = sum[t * 2] + sum[t * 2 + 1]; }
void down(int t, int l, int r) {
  sum[t * 2] *= l1[t];
  sum[t * 2 + 1] *= l1[t];
  l1[t * 2] *= l1[t];
  l1[t * 2 + 1] *= l1[t];
  l2[t * 2] *= l1[t];
  l2[t * 2 + 1] *= l1[t];
  l1[t] = 1;
  sum[t * 2 + 1] += (r - (l + r >> 1)) * l2[t];
  sum[t * 2] += ((l + r >> 1) - l + 1) * l2[t];
  sum[t] = sum[t * 2] + sum[t * 2 + 1];
  l2[t * 2] += l2[t];
  l2[t * 2 + 1] += l2[t];
  l2[t] = 0;
}
void build(int t, int l, int r) {
  L[t] = l;
  R[t] = r;
  l1[t] = 1;
  if (l == r) {
    sum[t] = a[l];
    return;
  }
  build(t * 2, l, l + r >> 1);
  build(t * 2 + 1, (l + r >> 1) + 1, r);
  up(t);
}
void upd(int t, double ql, double qr, double sl, double sr) {
  int l = L[t], r = R[t];
  if (r < ql || l > qr) return;
  if (ql <= l && r <= qr) {
    sum[t] *= 1.0 * (qr - ql) / (qr - ql + 1);
    ;
    sum[t] += 1.0 * s * (r - l + 1) / ((sr - sl + 1) * (qr - ql + 1));
    l1[t] *= 1.0 * (qr - ql) / (qr - ql + 1);
    l2[t] *= 1.0 * (qr - ql) / (qr - ql + 1);
    l2[t] += 1.0 * s / ((sr - sl + 1) * (qr - ql + 1));
    return;
  }
  down(t, l, r);
  upd(t * 2, ql, qr, sl, sr);
  upd(t * 2 + 1, ql, qr, sl, sr);
  up(t);
}
double qry(int t, int ql, int qr) {
  int l = L[t], r = R[t];
  if (r < ql || l > qr) return 0;
  if (ql <= l && r <= qr) return sum[t];
  down(t, l, r);
  return qry(t * 2, ql, qr) + qry(t * 2 + 1, ql, qr);
}
int main() {
  int n, q;
  scanf("%d", &n);
  scanf("%d", &q);
  for (int i = 1; i <= n; i++) scanf("%lf", a + i);
  build(1, 1, n);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 2) {
      int l, r;
      scanf("%d", &l);
      scanf("%d", &r);
      printf("%lf\n", qry(1, l, r));
    } else {
      double l1, r1, l2, r2;
      scanf("%lf%lf%lf%lf", &l1, &r1, &l2, &r2);
      double s1, s2;
      s1 = qry(1, l2, r2);
      s2 = qry(1, l1, r1);
      s = s1;
      upd(1, l1, r1, l2, r2);
      s = s2;
      upd(1, l2, r2, l1, r1);
    }
  }
}