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 = 100009;
const double eps = 1e-10;
struct sg_tree {
  double sum, mul, add;
} node[maxn << 2];
int n, m;
inline int sgn(double x) { return (x > -eps) - (x < eps); }
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
void update(int rt) {
  node[rt].sum = node[rt << 1].sum + node[rt << 1 | 1].sum;
}
void make_plus(int l, int r, int rt, double delta) {
  node[rt].sum += delta * (r - l + 1);
  node[rt].add += delta;
}
void make_mult(int l, int r, int rt, double delta) {
  node[rt].sum *= delta;
  node[rt].add *= delta;
  node[rt].mul *= delta;
}
void down(int l, int r, int rt) {
  int mid = (l + r) >> 1;
  if (sgn(node[rt].mul - 1)) {
    make_mult(l, mid, rt << 1, node[rt].mul);
    make_mult(mid + 1, r, rt << 1 | 1, node[rt].mul);
    node[rt].mul = 1;
  }
  if (sgn(node[rt].add)) {
    make_plus(l, mid, rt << 1, node[rt].add);
    make_plus(mid + 1, r, rt << 1 | 1, node[rt].add);
    node[rt].add = 0;
  }
}
void build(int l, int r, int rt) {
  node[rt].add = 0;
  node[rt].mul = 1;
  if (l == r) {
    node[rt].sum = read();
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  update(rt);
}
void modify(int l, int r, int rt, int left, int right, double delta) {
  if (l == left && right == r) {
    make_mult(l, r, rt, delta);
    return;
  }
  down(l, r, rt);
  int mid = (l + r) >> 1;
  if (right <= mid)
    modify(l, mid, rt << 1, left, right, delta);
  else if (left > mid)
    modify(mid + 1, r, rt << 1 | 1, left, right, delta);
  else
    modify(l, mid, rt << 1, left, mid, delta),
        modify(mid + 1, r, rt << 1 | 1, mid + 1, right, delta);
  update(rt);
}
void change(int l, int r, int rt, int left, int right, double delta) {
  if (l == left && right == r) {
    make_plus(l, r, rt, delta);
    return;
  }
  down(l, r, rt);
  int mid = (l + r) >> 1;
  if (right <= mid)
    change(l, mid, rt << 1, left, right, delta);
  else if (left > mid)
    change(mid + 1, r, rt << 1 | 1, left, right, delta);
  else
    change(l, mid, rt << 1, left, mid, delta),
        change(mid + 1, r, rt << 1 | 1, mid + 1, right, delta);
  update(rt);
}
double query(int l, int r, int rt, int left, int right) {
  if (l == left && right == r) return node[rt].sum;
  int mid = (l + r) >> 1;
  down(l, r, rt);
  if (right <= mid)
    return query(l, mid, rt << 1, left, right);
  else if (left > mid)
    return query(mid + 1, r, rt << 1 | 1, left, right);
  else
    return query(l, mid, rt << 1, left, mid) +
           query(mid + 1, r, rt << 1 | 1, mid + 1, right);
}
int main() {
  n = read();
  m = read();
  build(1, n, 1);
  while (m--) {
    int opt = read();
    if (opt == 1) {
      int l1 = read(), r1 = read(), l2 = read(), r2 = read(),
          len1 = r1 - l1 + 1, len2 = r2 - l2 + 1;
      double sum1 = query(1, n, 1, l1, r1), sum2 = query(1, n, 1, l2, r2);
      modify(1, n, 1, l1, r1, 1.0 * (len1 - 1) / len1);
      modify(1, n, 1, l2, r2, 1.0 * (len2 - 1) / len2);
      change(1, n, 1, l1, r1, sum2 / len1 / len2);
      change(1, n, 1, l2, r2, sum1 / len1 / len2);
    } else {
      int l = read(), r = read();
      printf("%lf\n", query(1, n, 1, l, r));
    }
  }
  return 0;
}