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 = 1000005;
const double eps = 1e-12;
const long long mod = 1000000009;
int n, Q;
int a[maxn];
double sg[maxn] = {0}, lza[maxn] = {0}, lzm[maxn] = {0};
void pushdown(int o, int l, int r) {
  if (lzm[o] != 1) {
    lzm[o << 1] *= lzm[o];
    lzm[o << 1 | 1] *= lzm[o];
    lza[o << 1] *= lzm[o];
    lza[o << 1 | 1] *= lzm[o];
    sg[o << 1] *= lzm[o];
    sg[o << 1 | 1] *= lzm[o];
    lzm[o] = 1;
  }
  if (lza[o] != 0) {
    lza[o << 1] += lza[o];
    lza[o << 1 | 1] += lza[o];
    int mid = (l + r) >> 1;
    sg[o << 1] += (mid - l) * lza[o];
    sg[o << 1 | 1] += (r - mid) * lza[o];
    lza[o] = 0;
  }
}
void pushup(int o) { sg[o] = sg[o << 1 | 1] + sg[o << 1]; }
void update(int o, int l, int r, int tl, int tr, int op, double x) {
  if (r <= tl || tr <= l) return;
  if (tl <= l && r <= tr) {
    if (op == 1)
      lzm[o] *= x, lza[o] *= x, sg[o] *= x;
    else
      lza[o] += x, sg[o] += (r - l) * x;
    return;
  }
  pushdown(o, l, r);
  int mid = (l + r) >> 1;
  update(o << 1, l, mid, tl, tr, op, x);
  update(o << 1 | 1, mid, r, tl, tr, op, x);
  pushup(o);
}
double query(int o, int l, int r, int tl, int tr) {
  if (r <= tl || tr <= l) return 0;
  if (tl <= l && r <= tr) return sg[o];
  pushdown(o, l, r);
  int mid = (l + r) >> 1;
  return query(o << 1, l, mid, tl, tr) + query(o << 1 | 1, mid, r, tl, tr);
}
int main() {
  int i, j;
  scanf("%d %d", &n, &Q);
  for (i = 0; i < n; i++) scanf("%d", a + i);
  for (i = 0; i < maxn; i++) lzm[i] = 1;
  for (i = 0; i < n; i++) update(1, 0, n, i, i + 1, 0, a[i]);
  while (Q--) {
    int t, l1, r1, l2, r2;
    scanf("%d %d %d", &t, &l1, &r1);
    if (t == 1) scanf("%d %d", &l2, &r2);
    --l1;
    --l2;
    if (t == 1) {
      double s1 = query(1, 0, n, l1, r1), s2 = query(1, 0, n, l2, r2);
      update(1, 0, n, l1, r1, 1, (r1 - l1 - 1.0) / (r1 - l1));
      update(1, 0, n, l1, r1, 0, s2 / (r1 - l1) / (r2 - l2));
      update(1, 0, n, l2, r2, 1, (r2 - l2 - 1.0) / (r2 - l2));
      update(1, 0, n, l2, r2, 0, s1 / (r1 - l1) / (r2 - l2));
    } else {
      printf("%.6lf\n", query(1, 0, n, l1, r1));
    }
  }
  return 0;
}