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 N = 510031;
int n, tests;
long double A[N], B[N];
int ar[N];
int tp;
pair<double, double> make_naive_query(int l, int r) {
  pair<double, double> res = make_pair(0, 0);
  for (int i = l; i <= r; i++) {
    res.first += A[i];
    res.second += B[i];
  }
  return res;
}
void make_naive_mult(int l, int r, double p) {
  for (int i = l; i <= r; i++) {
    B[i] *= p;
    A[i] *= p;
  }
}
void make_naive_add(int l, int r, double val, double chance) {
  for (int i = l; i <= r; i++) {
    A[i] += val;
    B[i] += chance;
  }
}
long double tA[N], tB[N], toaddA[N], toaddB[N], tomultA[N], tomultB[N];
void push(int v, int tl, int tr) {
  int sdfjahlfasfh = tl + tr;
  sdfjahlfasfh /= 2;
  tA[v * 2] *= tomultA[v];
  tA[v * 2] += toaddA[v] * (sdfjahlfasfh - tl + 1);
  toaddA[v * 2] *= tomultA[v];
  toaddA[v * 2] += toaddA[v];
  tomultA[v * 2] *= tomultA[v];
  tA[v * 2 + 1] *= tomultA[v];
  tA[v * 2 + 1] += toaddA[v] * (tr - sdfjahlfasfh);
  toaddA[v * 2 + 1] *= tomultA[v];
  toaddA[v * 2 + 1] += toaddA[v];
  tomultA[v * 2 + 1] *= tomultA[v];
  toaddA[v] = toaddB[v] = 0;
  tomultA[v] = tomultB[v] = 1;
}
void make_mult(int v, int tl, int tr, int l, int r, long double p) {
  if (l > r) return;
  if (tl == l && tr == r) {
    tA[v] *= p;
    tB[v] *= p;
    toaddA[v] *= p;
    tomultA[v] *= p;
    return;
  }
  push(v, tl, tr);
  int sdfjahlfasfh = tl + tr;
  sdfjahlfasfh /= 2;
  make_mult(v * 2, tl, sdfjahlfasfh, l, min(r, sdfjahlfasfh), p);
  make_mult(v * 2 + 1, sdfjahlfasfh + 1, tr, max(sdfjahlfasfh + 1, l), r, p);
  tA[v] = tA[v * 2] + tA[v * 2 + 1];
  tB[v] = tB[v * 2] + tB[v * 2 + 1];
}
void make_add(int v, int tl, int tr, int l, int r, long double p1,
              long double p2) {
  if (l > r) return;
  if (tl == l && tr == r) {
    tA[v] += p1 * (r - l + 1);
    tB[v] += p2 * (r - l + 1);
    toaddA[v] += p1;
    toaddB[v] += p2;
    return;
  }
  push(v, tl, tr);
  int sdfjahlfasfh = tl + tr;
  sdfjahlfasfh /= 2;
  make_add(v * 2, tl, sdfjahlfasfh, l, min(r, sdfjahlfasfh), p1, p2);
  make_add(v * 2 + 1, sdfjahlfasfh + 1, tr, max(sdfjahlfasfh + 1, l), r, p1,
           p2);
  tA[v] = tA[v * 2] + tA[v * 2 + 1];
  tB[v] = tB[v * 2] + tB[v * 2 + 1];
}
pair<long double, long double> make_query(int v, int tl, int tr, int l, int r) {
  if (l > r) return make_pair(0, 0);
  if (tl == l && tr == r) return make_pair(tA[v], tB[v]);
  push(v, tl, tr);
  int sdfjahlfasfh = tl + tr;
  sdfjahlfasfh /= 2;
  pair<long double, long double> p1 =
      make_query(v * 2, tl, sdfjahlfasfh, l, min(r, sdfjahlfasfh));
  pair<long double, long double> p2 =
      make_query(v * 2 + 1, sdfjahlfasfh + 1, tr, max(sdfjahlfasfh + 1, l), r);
  p1.first += p2.first;
  p1.second += p2.second;
  return p1;
}
void build(int v, int tl, int tr) {
  toaddA[v] = 0;
  toaddB[v] = 0;
  tomultA[v] = 1;
  tomultB[v] = 1;
  if (tl == tr) {
    tA[v] = ar[tl];
    tB[v] = 1.0;
    return;
  }
  int sdfjahlfasfh = tl + tr;
  sdfjahlfasfh /= 2;
  build(v * 2, tl, sdfjahlfasfh);
  build(v * 2 + 1, sdfjahlfasfh + 1, tr);
  tA[v] = tA[v * 2] + tA[v * 2 + 1];
  tB[v] = tB[v * 2] + tB[v * 2 + 1];
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> tests;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
  }
  for (int i = 1; i <= n; i++) {
    A[i] = ar[i];
    B[i] = 1.0;
  }
  build(1, 1, n);
  for (; tests; --tests) {
    cin >> tp;
    if (tp == 2) {
      int l, r;
      cin >> l >> r;
      pair<long double, long double> P = make_query(1, 1, n, l, r);
      cout.precision(7);
      cout << fixed << P.first << "\n";
    } else {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      pair<long double, long double> P1 = make_query(1, 1, n, l1, r1);
      pair<long double, long double> P2 = make_query(1, 1, n, l2, r2);
      long double ev1 = P1.first / (r1 - l1 + 1);
      long double ev2 = P2.first / (r2 - l2 + 1);
      make_mult(1, 1, n, l1, r1, 1.0 - 1.0 / (r1 - l1 + 1));
      make_mult(1, 1, n, l2, r2, 1.0 - 1.0 / (r2 - l2 + 1));
      make_add(1, 1, n, l1, r1, ev2 / (r1 - l1 + 1), 1.0 / (r1 - l1 + 1));
      make_add(1, 1, n, l2, r2, ev1 / (r2 - l2 + 1), 1.0 / (r2 - l2 + 1));
    }
  }
  return 0;
}