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>
const int N = 100010;
struct node {
  int l, r;
  double sum, add, mul;
} a[N << 2];
int n, q;
int read() {
  int d = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') d = d * 10 + c - 48, c = getchar();
  return d * f;
}
void pushup(int i) { a[i].sum = a[i << 1].sum + a[i << 1 | 1].sum; }
void pushdown(int i) {
  int i1 = i << 1, i2 = i1 | 1;
  a[i1].sum = a[i1].sum * a[i].mul + a[i].add * (a[i1].r - a[i1].l + 1);
  a[i1].mul *= a[i].mul;
  a[i1].add = a[i1].add * a[i].mul + a[i].add;
  a[i2].sum = a[i2].sum * a[i].mul + a[i].add * (a[i2].r - a[i2].l + 1);
  a[i2].mul *= a[i].mul;
  a[i2].add = a[i2].add * a[i].mul + a[i].add;
  a[i].add = 0;
  a[i].mul = 1;
}
void build(int i, int l, int r) {
  a[i].l = l;
  a[i].r = r;
  a[i].mul = 1;
  if (l == r) {
    a[i].sum = read();
    return;
  }
  int mid = (l + r) >> 1;
  build(i << 1, l, mid);
  build(i << 1 | 1, mid + 1, r);
  pushup(i);
}
double query(int i, int l, int r) {
  if (l <= a[i].l && a[i].r <= r) return a[i].sum;
  pushdown(i);
  int mid = (a[i].l + a[i].r) >> 1;
  if (r <= mid)
    return query(i << 1, l, r);
  else if (l > mid)
    return query(i << 1 | 1, l, r);
  else
    return query(i << 1, l, mid) + query(i << 1 | 1, mid + 1, r);
}
void Mul(int i, int l, int r, double x) {
  if (l <= a[i].l && a[i].r <= r) {
    a[i].sum *= x;
    a[i].add *= x;
    a[i].mul *= x;
    return;
  }
  pushdown(i);
  int mid = (a[i].l + a[i].r) >> 1;
  if (r <= mid)
    Mul(i << 1, l, r, x);
  else if (l > mid)
    Mul(i << 1 | 1, l, r, x);
  else
    Mul(i << 1, l, mid, x), Mul(i << 1 | 1, mid + 1, r, x);
  pushup(i);
}
void Add(int i, int l, int r, double x) {
  if (l <= a[i].l && a[i].r <= r) {
    a[i].sum += x * (a[i].r - a[i].l + 1);
    a[i].add += x;
    return;
  }
  pushdown(i);
  int mid = (a[i].l + a[i].r) >> 1;
  if (r <= mid)
    Add(i << 1, l, r, x);
  else if (l > mid)
    Add(i << 1 | 1, l, r, x);
  else
    Add(i << 1, l, mid, x), Add(i << 1 | 1, mid + 1, r, x);
  pushup(i);
}
int main() {
  n = read();
  q = read();
  build(1, 1, n);
  while (q--) {
    int op = read(), l = read(), r = read();
    if (op == 1) {
      int L = read(), R = read(), len1 = r - l + 1, len2 = R - L + 1;
      double sum1 = query(1, l, r), sum2 = query(1, L, R);
      Mul(1, l, r, (len1 - 1) * 1.0 / len1);
      Mul(1, L, R, (len2 - 1) * 1.0 / len2);
      Add(1, l, r, 1.0 / len1 * sum2 / len2);
      Add(1, L, R, 1.0 / len2 * sum1 / len1);
    } else
      printf("%.6lf\n", query(1, l, r));
  }
  return 0;
}