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 = (int)1e9 + 7;
const int FFTMOD = 1007681537;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
const int maxn = 1e5 + 5;
int n, q;
int a[maxn];
long double st[maxn << 2];
long double lz[maxn << 2][2];
void push(int p, int L, int R) {
  if (lz[p][0] != 1.0) {
    st[p] *= lz[p][0];
    if (L < R) {
      int t = p << 1;
      for (int i = (0); i < (2); i++) {
        if (lz[t + i][1]) {
          lz[t + i][1] *= lz[p][0];
        }
        lz[t + i][0] *= lz[p][0];
      }
    }
    lz[p][0] = 1.0;
  }
  if (lz[p][1] != 0.0) {
    st[p] += (R - L + 1) * lz[p][1];
    if (L < R) {
      int t = p << 1;
      for (int i = (0); i < (2); i++) {
        lz[t + i][1] += lz[p][1];
      }
    }
    lz[p][1] = 0.0;
  }
}
void pull(int p) { st[p] = st[p << 1] + st[p << 1 | 1]; }
void upd(int p, int l, int r, int L, int R, long double mul, long double add) {
  push(p, L, R);
  if (l > R || r < L) return;
  if (l <= L && r >= R) {
    lz[p][0] = mul, lz[p][1] = add;
    push(p, L, R);
    return;
  }
  upd(p << 1, l, r, L, L + R >> 1, mul, add);
  upd(p << 1 | 1, l, r, (L + R >> 1) + 1, R, mul, add);
  pull(p);
}
long double query(int p, int l, int r, int L, int R) {
  push(p, L, R);
  if (l > R || r < L) return 0.0;
  if (l <= L && r >= R) return st[p];
  return query(p << 1, l, r, L, L + R >> 1) +
         query(p << 1 | 1, l, r, (L + R >> 1) + 1, R);
}
void solve() {
  cin >> n >> q;
  for (int i = (0); i < (n << 2); i++) lz[i][0] = 1.0, lz[i][1] = 0.0;
  for (int i = (0); i < (n); i++) cin >> a[i], upd(1, i, i, 0, n - 1, 1, a[i]);
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int x[2], y[2];
      for (int i = (0); i < (2); i++) cin >> x[i] >> y[i], x[i]--, y[i]--;
      int len[2] = {y[0] - x[0] + 1, y[1] - x[1] + 1};
      long double t[2];
      for (int i = (0); i < (2); i++)
        t[i] = query(1, x[i ^ 1], y[i ^ 1], 0, n - 1) / len[i ^ 1];
      for (int i = (0); i < (2); i++) {
        upd(1, x[i], y[i], 0, n - 1, 1.0 * (len[i] - 1) / len[i], 0);
        upd(1, x[i], y[i], 0, n - 1, 1.0, t[i] / len[i]);
      }
    } else {
      int x, y;
      cin >> x >> y, x--, y--;
      cout << fixed << setprecision(9) << query(1, x, y, 0, n - 1) << "\n";
    }
  }
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  solve();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}