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 = 131072;
double s[maxn << 1], add[maxn << 1], mul[maxn << 1];
int N, Q;
double getsum(int l, int r, int L, int R, int rt) {
  if (L >= l && R <= r) return s[rt];
  if (add[rt]) {
    mul[rt << 1] *= mul[rt];
    add[rt << 1] *= mul[rt];
    add[rt << 1] += add[rt] / 2;
    mul[rt << 1 | 1] *= mul[rt];
    add[rt << 1 | 1] *= mul[rt];
    add[rt << 1 | 1] += add[rt] / 2;
    s[rt << 1] *= mul[rt];
    s[rt << 1] += add[rt] / 2;
    s[rt << 1 | 1] *= mul[rt];
    s[rt << 1 | 1] += add[rt] / 2;
    mul[rt] = 1;
    add[rt] = 0;
  }
  int mid = L + R >> 1;
  double res = 0;
  if (mid >= l) res = getsum(l, r, L, mid, rt << 1);
  if (mid < r) res += getsum(l, r, mid + 1, R, rt << 1 | 1);
  return res;
}
void modify(int l, int r, double upd, int L, int R, int rt) {
  if (L >= l && R <= r) {
    mul[rt] *= 1.0 * (r - l) / (r - l + 1);
    add[rt] *= 1.0 * (r - l) / (r - l + 1);
    add[rt] += upd * (R - L + 1) / (r - l + 1);
    s[rt] = upd / (r - l + 1) * (R - L + 1) + s[rt] / (r - l + 1) * (r - l);
    return;
  }
  if (add[rt]) {
    mul[rt << 1] *= mul[rt];
    add[rt << 1] *= mul[rt];
    add[rt << 1] += add[rt] / 2;
    mul[rt << 1 | 1] *= mul[rt];
    add[rt << 1 | 1] *= mul[rt];
    add[rt << 1 | 1] += add[rt] / 2;
    s[rt << 1] *= mul[rt];
    s[rt << 1] += add[rt] / 2;
    s[rt << 1 | 1] *= mul[rt];
    s[rt << 1 | 1] += add[rt] / 2;
    mul[rt] = 1;
    add[rt] = 0;
  }
  int mid = L + R >> 1;
  if (mid >= l) modify(l, r, upd, L, mid, rt << 1);
  if (mid < r) modify(l, r, upd, mid + 1, R, rt << 1 | 1);
  s[rt] = s[rt << 1] + s[rt << 1 | 1];
}
int main() {
  scanf("%d%d", &N, &Q);
  for (int i = 1; i <= N; i++) scanf("%lf", s + maxn + i - 1);
  for (int i = maxn - 1; i; i--) s[i] = s[i << 1] + s[i << 1 | 1];
  for (int i = 1; i < maxn + maxn; i++) mul[i] = 1;
  for (int i = 1; i <= Q; i++) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l1, r1, l2, r2;
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double t1 = getsum(l1, r1, 1, maxn, 1), t2 = getsum(l2, r2, 1, maxn, 1);
      modify(l1, r1, t2 / (r2 - l2 + 1), 1, maxn, 1);
      modify(l2, r2, t1 / (r1 - l1 + 1), 1, maxn, 1);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%lf\n", getsum(l, r, 1, maxn, 1));
    }
  }
}