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, M = 1000010, P = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int INF = 0xcfcfcfcf;
const double eps = 1e-9, pi = asin(1) * 2;
int ADD(int a, int b) { return a + b >= P ? a + b - P : a + b; }
int MINUS(int a, int b) { return a - b < 0 ? a - b + P : a - b; }
int n, q;
int a[N];
bool equal(double a, double b) { return abs(a - b) <= eps; }
namespace SEGMENT_TREE {
double sum[N << 4], tag1[N << 4], tag2[N << 4];
void pushup(int p) { sum[p] = sum[(p << 1)] + sum[(p << 1 | 1)]; }
void build(int p, int l, int r) {
  tag1[p] = 1, tag2[p] = 0;
  if (l == r) {
    sum[p] = a[l];
    return;
  }
  int mid = l + r >> 1;
  build((p << 1), l, mid);
  build((p << 1 | 1), mid + 1, r);
  pushup(p);
}
void update1(int p, double v) { sum[p] *= v, tag1[p] *= v, tag2[p] *= v; }
void update2(int p, int len, double v) { sum[p] += len * v, tag2[p] += v; }
void pushdown(int p, int l, int r) {
  int mid = l + r >> 1;
  if (!equal(tag1[p], 1)) {
    update1((p << 1), tag1[p]);
    update1((p << 1 | 1), tag1[p]);
    tag1[p] = 1;
  }
  if (!equal(tag2[p], 0)) {
    update2((p << 1), mid - l + 1, tag2[p]);
    update2((p << 1 | 1), r - mid, tag2[p]);
    tag2[p] = 0;
  }
}
void update1(int p, int l, int r, int ql, int qr, double v) {
  if (ql <= l && r <= qr) return update1(p, v);
  pushdown(p, l, r);
  int mid = l + r >> 1;
  if (ql <= mid) update1((p << 1), l, mid, ql, qr, v);
  if (qr > mid) update1((p << 1 | 1), mid + 1, r, ql, qr, v);
  pushup(p);
}
void update2(int p, int l, int r, int ql, int qr, double v) {
  if (ql <= l && r <= qr) return update2(p, r - l + 1, v);
  pushdown(p, l, r);
  int mid = l + r >> 1;
  if (ql <= mid) update2((p << 1), l, mid, ql, qr, v);
  if (qr > mid) update2((p << 1 | 1), mid + 1, r, ql, qr, v);
  pushup(p);
}
double query(int p, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return sum[p];
  pushdown(p, l, r);
  int mid = l + r >> 1;
  double ans = 0;
  if (ql <= mid) ans += query((p << 1), l, mid, ql, qr);
  if (qr > mid) ans += query((p << 1 | 1), mid + 1, r, ql, qr);
  return ans;
}
}  // namespace SEGMENT_TREE
using SEGMENT_TREE::build;
using SEGMENT_TREE::query;
using SEGMENT_TREE::update1;
using SEGMENT_TREE::update2;
namespace IO {
long long read() {
  long long s = 0;
  bool flag = false;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') flag = true;
  for (; '0' <= ch && ch <= '9'; ch = getchar())
    s = (s << 3) + (s << 1) + (ch ^ '0');
  if (flag) return -s;
  return s;
}
}  // namespace IO
using IO::read;
int main() {
  n = read(), q = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    int opt = read();
    if (opt == 1) {
      int l1 = read(), r1 = read(), l2 = read(), r2 = read();
      int L1 = r1 - l1 + 1, L2 = r2 - l2 + 1;
      double E1 = query(1, 1, n, l1, r1), E2 = query(1, 1, n, l2, r2);
      update1(1, 1, n, l1, r1, 1.0 * (L1 - 1) / L1);
      update1(1, 1, n, l2, r2, 1.0 * (L2 - 1) / L2);
      update2(1, 1, n, l1, r1, E2 / L1 / L2);
      update2(1, 1, n, l2, r2, E1 / L1 / L2);
    }
    if (opt == 2) {
      int l = read(), r = read();
      printf("%.7lf\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}