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 = 1e5 + 10;
const double eps = 1e-8;
double W, w;
struct Node {
  double sum, fl, delta;
  int sz;
};
Node pool[maxn << 2];
void push_up(int u) {
  int ls = u << 1, rs = ls | 1;
  pool[u].sum = pool[ls].sum + pool[rs].sum;
}
void push_down(int u) {
  if (pool[u].delta > eps) {
    int ls = u << 1, rs = ls | 1;
    double dl = pool[u].delta, fl = pool[u].fl;
    pool[u].delta = 0.0;
    pool[u].fl = 1.0;
    pool[ls].sum = pool[ls].sum * fl + dl * pool[ls].sz;
    pool[rs].sum = pool[rs].sum * fl + dl * pool[rs].sz;
    pool[ls].delta = pool[ls].delta * fl + dl;
    pool[rs].delta = pool[rs].delta * fl + dl;
    pool[ls].fl *= fl;
    pool[rs].fl *= fl;
  }
}
void build(int u, int L, int R) {
  pool[u].fl = 1.0;
  pool[u].delta = 0.0;
  if (L == R) {
    scanf("%lf", &W);
    pool[u].sum = 1.0 * W;
    pool[u].sz = 1;
    return;
  }
  int md = L + R >> 1, ls = u << 1, rs = ls | 1;
  build(ls, L, md);
  build(rs, md + 1, R);
  push_up(u);
  pool[u].sz = pool[ls].sz + pool[rs].sz;
}
void update(int u, int L, int R, int l, int r, double fl, double dl) {
  if (l <= L and R <= r) {
    pool[u].sum = pool[u].sum * fl + dl * pool[u].sz;
    pool[u].fl *= fl;
    pool[u].delta = pool[u].delta * fl + dl;
    return;
  }
  int md = L + R >> 1, ls = u << 1, rs = ls | 1;
  push_down(u);
  if (r <= md)
    update(ls, L, md, l, r, fl, dl);
  else if (md < l)
    update(rs, md + 1, R, l, r, fl, dl);
  else
    update(ls, L, md, l, r, fl, dl), update(rs, md + 1, R, l, r, fl, dl);
  push_up(u);
}
double calc(int u, int L, int R, int l, int r) {
  if (l <= L and R <= r) return pool[u].sum;
  push_down(u);
  int md = L + R >> 1, ls = u << 1, rs = ls | 1;
  if (r <= md)
    return calc(ls, L, md, l, r);
  else if (md < l)
    return calc(rs, md + 1, R, l, r);
  else
    return calc(ls, L, md, l, r) + calc(rs, md + 1, R, l, r);
}
int n, m, l1, r1, l2, r2, tp;
double f1, f2, d1, d2;
int main() {
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  while (m--) {
    scanf("%d", &tp);
    if (tp == 1) {
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      W = calc(1, 1, n, l1, r1);
      w = calc(1, 1, n, l2, r2);
      f1 = 1.0 / (r1 - l1 + 1);
      f2 = 1.0 / (r2 - l2 + 1);
      d1 = w * f1 * f2;
      d2 = W * f1 * f2;
      f1 = f1 * (r1 - l1);
      f2 = f2 * (r2 - l2);
      update(1, 1, n, l1, r1, f1, d1);
      update(1, 1, n, l2, r2, f2, d2);
    } else {
      scanf("%d%d", &l1, &r1);
      printf("%.12lf\n", calc(1, 1, n, l1, r1));
    }
  }
  return 0;
}