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 mod = 1e9 + 7;
template <typename T1, typename T2>
inline void chkmin(T1 &x, T2 y) {
  if (y < x) x = y;
};
template <typename T1, typename T2>
inline void chkmax(T1 &x, T2 y) {
  if (y > x) x = y;
};
int gcd(int x, int y) { return (y == 0 ? x : gcd(y, x % y)); }
inline int mul(int a, int b) { return (1ll * a * b) % mod; }
inline int sub(int a, int b) {
  a -= b;
  if (a < 0) a += mod;
  if (a >= mod) a -= mod;
  return a;
}
inline int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
  return a;
}
inline int expo(int a, int b) {
  int values = 1;
  while (b) {
    if (b & 1) values = mul(values, a);
    a = mul(a, a);
    b >>= 1;
  }
  return values % mod;
}
const int MAXN = 1e6 + 100;
int n, q;
double a[MAXN], ad[MAXN], mult[MAXN], t[MAXN];
void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = a[tl];
    mult[v] = 1;
    return;
  }
  int md = (tl + tr) >> 1;
  build(2 * v, tl, md);
  build(2 * v + 1, md + 1, tr);
  t[v] = t[2 * v] + t[2 * v + 1];
  mult[v] = 1;
}
void call(int v, int tl, int tr) {
  t[2 * v] *= mult[v];
  t[2 * v + 1] *= mult[v];
  mult[2 * v] *= mult[v];
  mult[2 * v + 1] *= mult[v];
  ad[2 * v] *= mult[v];
  ad[2 * v + 1] *= mult[v];
  mult[v] = 1;
  int md = (tl + tr) >> 1;
  t[2 * v] += (double)(md - tl + 1) * ad[v];
  t[2 * v + 1] += (double)(tr - md) * ad[v];
  ad[2 * v] += ad[v];
  ad[2 * v + 1] += ad[v];
  ad[v] = 0;
  t[v] = t[2 * v] + t[2 * v + 1];
}
void update(int v, int tl, int tr, int l, int r, double ok, double ko) {
  if (l > r) return;
  if (tl == l and tr == r) {
    t[v] *= ok;
    t[v] += (double)(tr - tl + 1) * ko;
    mult[v] *= ok;
    ad[v] *= ok;
    ad[v] += ko;
    return;
  }
  call(v, tl, tr);
  int md = (tl + tr) >> 1;
  update(2 * v, tl, md, l, min(md, r), ok, ko);
  update(2 * v + 1, md + 1, tr, max(l, md + 1), r, ok, ko);
  t[v] = t[2 * v] + t[2 * v + 1];
  return;
}
double query(int v, int tl, int tr, int l, int r) {
  if (tl == l and tr == r) return t[v];
  if (l > r) return 0;
  call(v, tl, tr);
  int md = (tl + tr) >> 1;
  return (query(2 * v, tl, md, l, min(r, md)) +
          query(2 * v + 1, md + 1, tr, max(md + 1, l), r));
}
void solve() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  build(1, 0, n - 1);
  while (q--) {
    int type, l, r;
    cin >> type >> l >> r;
    l--;
    r--;
    if (type == 1) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      double res = query(1, 0, n - 1, l, r) / double(r - l + 1);
      double val = query(1, 0, n - 1, u, v) / double(v - u + 1);
      update(1, 0, n - 1, l, r, double(r - l) / double(r - l + 1),
             val / double(r - l + 1));
      update(1, 0, n - 1, u, v, double(v - u) / double(v - u + 1),
             res / double(v - u + 1));
    } else
      cout << fixed << setprecision(7) << query(1, 0, n - 1, l, r) << endl;
  }
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}