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 t[800000];
double add[800000];
double mult[800000];
int n, q;
int a[100000];
void push(int v, int tl, int tr) {
  t[v * 2] *= mult[v];
  t[v * 2 + 1] *= mult[v];
  mult[v * 2] *= mult[v];
  mult[v * 2 + 1] *= mult[v];
  add[v * 2] *= mult[v];
  add[v * 2 + 1] *= mult[v];
  mult[v] = 1;
  int tm = (tl + tr) / 2;
  t[v * 2] += (double)(tm - tl + 1) * add[v];
  t[v * 2 + 1] += (double)(tr - tm) * add[v];
  add[v * 2] += add[v];
  add[v * 2 + 1] += add[v];
  add[v] = 0;
  t[v] = t[v * 2] + t[v * 2 + 1];
}
void build(int v, int tl, int tr) {
  if (tl > tr) {
    return;
  }
  if (tl == tr) {
    t[v] = a[tl];
    return;
  }
  int tm = (tl + tr) / 2;
  build(v * 2, tl, tm);
  build(v * 2 + 1, tm + 1, tr);
  t[v] = t[v * 2] + t[v * 2 + 1];
}
void update(int v, int tl, int tr, int l, int r, double mult_val,
            double add_val) {
  if (l > r) return;
  if (tl == l && tr == r) {
    t[v] *= mult_val;
    t[v] += double(tr - tl + 1) * add_val;
    mult[v] *= mult_val;
    add[v] *= mult_val;
    add[v] += add_val;
    return;
  }
  push(v, tl, tr);
  int tm = (tl + tr) / 2;
  update(v * 2, tl, tm, l, min(tm, r), mult_val, add_val);
  update(v * 2 + 1, tm + 1, tr, max(tm + 1, l), r, mult_val, add_val);
  t[v] = t[2 * v] + t[2 * v + 1];
}
double sum(int v, int tl, int tr, int l, int r) {
  if (l > r) {
    return 0;
  }
  if (l == tl && r == tr) {
    return t[v];
  }
  push(v, tl, tr);
  int tm = (tl + tr) / 2;
  return sum(v * 2, tl, tm, l, min(r, tm)) +
         sum(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
}
int main() {
  for (int i = 0; i < 400000; i++) mult[i] = 1;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  build(1, 0, n - 1);
  for (int i = 0; i < q; i++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l1, r1, l2, r2;
      scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
      l1--;
      r1--;
      l2--;
      r2--;
      double mean1 = sum(1, 0, n - 1, l1, r1) / (r1 - l1 + 1);
      double mean2 = sum(1, 0, n - 1, l2, r2) / (r2 - l2 + 1);
      double koef1 = (r1 - l1) / (r1 - l1 + 1.);
      double koef2 = (r2 - l2) / (r2 - l2 + 1.);
      double smallKoef1 = 1. / (r1 - l1 + 1);
      double smallKoef2 = 1. / (r2 - l2 + 1);
      update(1, 0, n - 1, l1, r1, koef1, smallKoef1 * mean2);
      update(1, 0, n - 1, l2, r2, koef2, smallKoef2 * mean1);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      l--;
      r--;
      printf("%1.4lf\n", sum(1, 0, n - 1, l, r));
    }
  }
}