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;
int n;
double tr[4 * N], lzs[4 * N], lzm[4 * N];
long long v[N];
void build(int no, int l, int r) {
  lzs[no] = 0;
  lzm[no] = 1.0;
  if (l == r) {
    tr[no] = v[l];
    return;
  }
  int nxt = (no << 1), mid = (l + r) >> 1;
  build(nxt, l, mid);
  build(nxt + 1, mid + 1, r);
  tr[no] = tr[nxt] + tr[nxt + 1];
}
void propagate(int no, int l, int r) {
  tr[no] *= lzm[no];
  tr[no] += lzs[no] * (double)(r - l + 1);
  if (l != r) {
    int nxt = (no << 1);
    lzm[nxt] *= lzm[no];
    lzs[nxt] *= lzm[no];
    lzs[nxt] += lzs[no];
    lzm[nxt + 1] *= lzm[no];
    lzs[nxt + 1] *= lzm[no];
    lzs[nxt + 1] += lzs[no];
  }
  lzm[no] = 1.0;
  lzs[no] = 0.0;
}
double query(int no, int l, int r, int i, int j) {
  propagate(no, l, r);
  if (r < i || l > j) return 0;
  if (l >= i && r <= j) return tr[no];
  int nxt = (no << 1), mid = (l + r) >> 1;
  return query(nxt, l, mid, i, j) + query(nxt + 1, mid + 1, r, i, j);
}
void update(int no, int l, int r, int i, int j, double m, double x) {
  propagate(no, l, r);
  if (r < i || l > j) return;
  if (l >= i && r <= j) {
    lzm[no] = m;
    lzs[no] = x;
    propagate(no, l, r);
    return;
  }
  int nxt = (no << 1), mid = (l + r) >> 1;
  update(nxt, l, mid, i, j, m, x);
  update(nxt + 1, mid + 1, r, i, j, m, x);
  tr[no] = tr[nxt] + tr[nxt + 1];
}
inline void go(int l1, int r1, int l2, int r2) {
  double sa = query(1, 1, n, l1, r1);
  double sb = query(1, 1, n, l2, r2);
  double fooA = sa / (r1 - l1 + 1);
  double fooB = sb / (r2 - l2 + 1);
  update(1, 1, n, l1, r1, (double)(r1 - l1) / (double)(r1 - l1 + 1),
         fooB / (r1 - l1 + 1));
  update(1, 1, n, l2, r2, (double)(r2 - l2) / (double)(r2 - l2 + 1),
         fooA / (r2 - l2 + 1));
}
int q;
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", v + i);
  }
  build(1, 1, n);
  while (q--) {
    int op, l1, r1, l2, r2;
    scanf("%d %d %d", &op, &l1, &r1);
    if (op == 1) {
      scanf("%d %d", &l2, &r2);
      go(l1, r1, l2, r2);
    } else {
      printf("%.10lf\n", query(1, 1, n, l1, r1));
    }
  }
  return 0;
}