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 = 1e5 + 100;
double sum[N << 2], lazy[N << 2], lazy2[N << 2];
int n, m;
void pushdown(int l, int r, int rt) {
  lazy[rt << 1] *= lazy[rt];
  lazy2[rt << 1] *= lazy[rt];
  lazy2[rt << 1] += lazy2[rt];
  lazy[rt << 1 | 1] *= lazy[rt];
  lazy2[rt << 1 | 1] *= lazy[rt];
  lazy2[rt << 1 | 1] += lazy2[rt];
  sum[rt << 1] *= lazy[rt];
  sum[rt << 1] += lazy2[rt] * l;
  sum[rt << 1 | 1] *= lazy[rt];
  sum[rt << 1 | 1] += lazy2[rt] * r;
  lazy[rt] = 1;
  lazy2[rt] = 0;
}
void pushup(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; }
void build(int l, int r, int rt) {
  lazy[rt] = 1.0;
  lazy2[rt] = 0;
  if (l == r) {
    scanf("%lf", &sum[rt]);
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  pushup(rt);
}
double query(int L, int R, int l, int r, int rt) {
  if (L <= l && R >= r) {
    return sum[rt];
  }
  double ret = 0;
  int mid = r + l >> 1;
  pushdown(mid - l + 1, r - mid, rt);
  if (mid >= L) ret += query(L, R, l, mid, rt << 1);
  if (mid < R) ret += query(L, R, mid + 1, r, rt << 1 | 1);
  pushup(rt);
  return ret;
}
void update(int L, int R, double x, double y, int l, int r, int rt) {
  if (L <= l && R >= r) {
    lazy[rt] *= x;
    lazy2[rt] *= x;
    lazy2[rt] += y;
    sum[rt] = sum[rt] * x + y * (r - l + 1);
    return;
  }
  int mid = r + l >> 1;
  pushdown(l - mid + 1, r - mid, rt);
  if (mid >= L) update(L, R, x, y, l, mid, rt << 1);
  if (mid < R) update(L, R, x, y, mid + 1, r, rt << 1 | 1);
  pushup(rt);
}
int main() {
  cin >> n >> m;
  build(1, n, 1);
  for (int i = 1; i <= m; i++) {
    int op, l, r, l2, r2;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d %d %d %d", &l, &r, &l2, &r2);
      double sum1 = query(l, r, 1, n, 1);
      double sum2 = query(l2, r2, 1, n, 1);
      int x = (r - l + 1), y = (r2 - l2 + 1);
      update(l, r, (x - 1) * 1.0 / x, sum2 / x / y, 1, n, 1);
      update(l2, r2, (y - 1) * 1.0 / y, sum1 / x / y, 1, n, 1);
    } else {
      scanf("%d %d", &l, &r);
      printf("%lf\n", query(l, r, 1, n, 1));
    }
  }
  return 0;
}