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;
double a[200007];
struct node {
  double v, mul, add;
} st[200000 * 5];
int sgn(double a, double b) { return fabs(a - b) < 1e-10; }
void bt(int root, int l, int r) {
  st[root].mul = 1;
  st[root].add = 0;
  if (l == r) {
    st[root].v = a[l];
  } else {
    int m = (l + r) / 2;
    bt(root * 2, l, m);
    bt(root * 2 + 1, m + 1, r);
    st[root].v = st[root * 2].v + st[root * 2 + 1].v;
  }
  return;
}
void pushdown(int root, int l, int r) {
  int m = (l + r) / 2;
  st[root * 2].v = (st[root * 2].v * st[root].mul + st[root].add * (m - l + 1));
  st[root * 2 + 1].v =
      (st[root * 2 + 1].v * st[root].mul + st[root].add * (r - m));
  st[root * 2].mul = (st[root * 2].mul * st[root].mul);
  st[root * 2 + 1].mul = (st[root * 2 + 1].mul * st[root].mul);
  st[root * 2].add = (st[root * 2].add * st[root].mul + st[root].add);
  st[root * 2 + 1].add = (st[root * 2 + 1].add * st[root].mul + st[root].add);
  st[root].mul = 1;
  st[root].add = 0;
  return;
}
void ud1(int root, int stdl, int stdr, int l, int r, double k) {
  if (r < stdl || stdr < l) {
    return;
  }
  if (l <= stdl && stdr <= r) {
    st[root].v = (st[root].v * k);
    st[root].mul = (st[root].mul * k);
    st[root].add = (st[root].add * k);
    return;
  }
  pushdown(root, stdl, stdr);
  int m = (stdl + stdr) / 2;
  ud1(root * 2, stdl, m, l, r, k);
  ud1(root * 2 + 1, m + 1, stdr, l, r, k);
  st[root].v = (st[root * 2].v + st[root * 2 + 1].v);
  return;
}
void ud2(int root, int stdl, int stdr, int l, int r, double k) {
  if (r < stdl || stdr < l) {
    return;
  }
  if (l <= stdl && stdr <= r) {
    st[root].add = (st[root].add + k);
    st[root].v = (st[root].v + k * (stdr - stdl + 1));
    return;
  }
  pushdown(root, stdl, stdr);
  int m = (stdl + stdr) / 2;
  ud2(root * 2, stdl, m, l, r, k);
  ud2(root * 2 + 1, m + 1, stdr, l, r, k);
  st[root].v = (st[root * 2].v + st[root * 2 + 1].v);
  return;
}
double query(int root, int stdl, int stdr, int l, int r) {
  if (r < stdl || stdr < l) {
    return 0;
  }
  if (l <= stdl && stdr <= r) {
    return st[root].v;
  }
  pushdown(root, stdl, stdr);
  int m = (stdl + stdr) / 2;
  return (query(root * 2, stdl, m, l, r) +
          query(root * 2 + 1, m + 1, stdr, l, r));
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%lf", &a[i]);
  }
  bt(1, 1, n);
  while (m--) {
    int chk;
    scanf("%d", &chk);
    int x, y;
    if (chk == 1) {
      int l1, r1, l2, r2;
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double s1 = query(1, 1, n, l1, r1);
      double s2 = query(1, 1, n, l2, r2);
      double ave1 = s1 / (r1 - l1 + 1);
      double ave2 = s2 / (r2 - l2 + 1);
      int len1 = r1 - l1 + 1;
      int len2 = r2 - l2 + 1;
      ud1(1, 1, n, l1, r1, (len1 - 1) * 1.0 / len1);
      ud2(1, 1, n, l1, r1, ave2 * 1.0 / len1);
      ud1(1, 1, n, l2, r2, (len2 - 1) * 1.0 / len2);
      ud2(1, 1, n, l2, r2, ave1 * 1.0 / len2);
    } else if (chk == 2) {
      int l1, r1;
      scanf("%d%d", &l1, &r1);
      printf("%.10f\n", query(1, 1, n, l1, r1));
    }
  }
  getchar();
  getchar();
  return 0;
}