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 long double PI = acos(-1);
const int INF = 2e9 + 5;
const long long LINF = 1e18 + 5;
const long long MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
int n;
double a[MAXN];
double lp[4 * MAXN];
double sp[4 * MAXN];
double t[4 * MAXN];
int lazy[4 * MAXN];
void build(int nodo, int l, int r) {
  sp[nodo] = 0.0;
  lp[nodo] = 1.0;
  lazy[nodo] = 0;
  if (r - l < 2) {
    t[nodo] = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(nodo << 1, l, mid);
  build(nodo << 1 | 1, mid, r);
  t[nodo] = t[nodo << 1] + t[nodo << 1 | 1];
}
void update(int nodo, int a, int b, int l, int r, double f, double s) {
  if (lazy[nodo]) {
    t[nodo] = lp[nodo] * t[nodo] + sp[nodo] * (b - a);
    if (b - a > 1) {
      lazy[nodo << 1] = 1;
      lazy[nodo << 1 | 1] = 1;
      lp[nodo << 1] = lp[nodo << 1] * lp[nodo];
      lp[nodo << 1 | 1] = lp[nodo << 1 | 1] * lp[nodo];
      sp[nodo << 1] = sp[nodo << 1] * lp[nodo] + sp[nodo];
      sp[nodo << 1 | 1] = sp[nodo << 1 | 1] * lp[nodo] + sp[nodo];
    }
    lazy[nodo] = 0;
    sp[nodo] = 0.0;
    lp[nodo] = 1.0;
  }
  if (a >= b || r <= a || l >= b) return;
  if (l <= a && b <= r) {
    t[nodo] = f * t[nodo] + s * (b - a);
    if (b - a > 1) {
      lazy[nodo << 1] = 1;
      lazy[nodo << 1 | 1] = 1;
      lp[nodo << 1] = lp[nodo << 1] * f;
      lp[nodo << 1 | 1] = lp[nodo << 1 | 1] * f;
      sp[nodo << 1] = sp[nodo << 1] * f + s;
      sp[nodo << 1 | 1] = sp[nodo << 1 | 1] * f + s;
    }
    return;
  }
  int mid = (a + b) >> 1;
  update(nodo << 1, a, mid, l, r, f, s);
  update(nodo << 1 | 1, mid, b, l, r, f, s);
  t[nodo] = t[nodo << 1] + t[nodo << 1 | 1];
}
double Query(int nodo, int a, int b, int l, int r) {
  if (lazy[nodo]) {
    t[nodo] = lp[nodo] * t[nodo] + sp[nodo] * (b - a);
    if (b - a > 1) {
      lazy[nodo << 1] = 1;
      lazy[nodo << 1 | 1] = 1;
      lp[nodo << 1] = lp[nodo << 1] * lp[nodo];
      lp[nodo << 1 | 1] = lp[nodo << 1 | 1] * lp[nodo];
      sp[nodo << 1] = sp[nodo << 1] * lp[nodo] + sp[nodo];
      sp[nodo << 1 | 1] = sp[nodo << 1 | 1] * lp[nodo] + sp[nodo];
    }
    lazy[nodo] = 0;
    sp[nodo] = 0.0;
    lp[nodo] = 1.0;
  }
  if (a >= b || r <= a || l >= b) return 0;
  if (l <= a && b <= r) {
    return t[nodo];
  }
  int mid = (a + b) >> 1;
  double q1 = Query(nodo << 1, a, mid, l, r);
  double q2 = Query(nodo << 1 | 1, mid, b, l, r);
  return q1 + q2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int Q;
  cin >> n >> Q;
  for (int i = 0; i < n; ++i) cin >> a[i];
  build(1, 0, n);
  int tipo;
  int l1, l2, r1, r2;
  int l, r;
  for (int q = 0; q < Q; ++q) {
    cin >> tipo;
    if (tipo == 1) {
      cin >> l1 >> r1 >> l2 >> r2;
      --l1, --l2;
      int n1 = r1 - l1;
      int n2 = r2 - l2;
      double v1 = Query(1, 0, n, l1, r1);
      double v2 = Query(1, 0, n, l2, r2);
      v1 /= n1;
      v2 /= n2;
      double f2 = (n2 - 1.0) / (1.0 * n2);
      double f1 = (n1 - 1.0) / (1.0 * n1);
      update(1, 0, n, l2, r2, f2, v1 / n2);
      update(1, 0, n, l1, r1, f1, v2 / n1);
    } else {
      cin >> l >> r;
      --l;
      cout << fixed << setprecision(12) << Query(1, 0, n, l, r) << endl;
    }
  }
  return 0;
}