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;
int N, Q;
int a[100010];
struct Seg {
  double val, a, b;
  int len;
} seg[100010 << 2];
void build(int id, int l, int r) {
  seg[id].len = r - l + 1;
  seg[id].a = 1;
  seg[id].b = 0;
  if (l == r) {
    seg[id].val = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(id << 1, l, mid);
  build((id << 1) | 1, mid + 1, r);
  seg[id].val = seg[id << 1].val + seg[(id << 1) | 1].val;
}
void modify(int id, double a, double b) {
  seg[id].a *= a;
  seg[id].b = seg[id].b * a + b;
  seg[id].val = seg[id].val * a + b * seg[id].len;
}
void down(int id) {
  modify(id << 1, seg[id].a, seg[id].b);
  modify((id << 1) | 1, seg[id].a, seg[id].b);
  seg[id].a = 1, seg[id].b = 0;
}
void update(int id, int l, int r, int ul, int ur, double a, double b) {
  if (ul > r || ur < l) return;
  if (ul <= l && r <= ur) {
    modify(id, a, b);
    return;
  }
  int mid = (l + r) >> 1;
  update(id << 1, l, mid, ul, ur, a, b);
  update((id << 1) | 1, mid + 1, r, ul, ur, a, b);
  seg[id].val = seg[id << 1].val + seg[(id << 1) | 1].val;
}
double query(int id, int l, int r, int ql, int qr) {
  if (ql > r || qr < l) return 0;
  if (ql <= l && r <= qr) return seg[id].val;
  down(id);
  int mid = (l + r) >> 1;
  return query(id << 1, l, mid, ql, qr) +
         query((id << 1) | 1, mid + 1, r, ql, qr);
}
int main() {
  scanf("%d%d", &N, &Q);
  for (int i = 0; i < N; i++) scanf("%d", &a[i]);
  build(1, 0, N - 1);
  while (Q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l1, r1, l2, r2;
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      l1--, r1--, l2--, r2--;
      double d1 = r1 - l1 + 1, d2 = r2 - l2 + 1;
      double q1 = query(1, 0, N - 1, l1, r1), q2 = query(1, 0, N - 1, l2, r2);
      update(1, 0, N - 1, l1, r1, (double)(d1 - 1) / d1, (double)q2 / d1 / d2);
      update(1, 0, N - 1, l2, r2, (double)(d2 - 1) / d2, (double)q1 / d1 / d2);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      l--, r--;
      printf("%.7f\n", query(1, 0, N - 1, l, r));
    }
  }
  return 0;
}