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 long maxn = 1e5 + 500;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 500;
long double lazyz[maxn * 8];
long double lazys[maxn * 8];
long double sum[maxn * 8];
void shift(long long node, long long l, long long r) {
  long long mid = (l + r) / 2;
  lazyz[2 * node] *= lazyz[node];
  lazys[2 * node] *= lazyz[node];
  sum[2 * node] *= lazyz[node];
  lazyz[2 * node + 1] *= lazyz[node];
  lazys[2 * node + 1] *= lazyz[node];
  sum[2 * node + 1] *= lazyz[node];
  lazyz[node] = 1;
  lazys[2 * node] += lazys[node];
  sum[2 * node] += (mid - l) * lazys[node];
  lazys[2 * node + 1] += lazys[node];
  sum[2 * node + 1] += (r - mid) * lazys[node];
  lazys[node] = 0;
}
void updatez(long long L, long long R, long long node, long long l, long long r,
             long double z) {
  if (l <= L && R <= r) {
    shift(node, L, R);
    lazyz[node] *= z;
    sum[node] *= z;
    return;
  }
  if (R <= l || r <= L) {
    return;
  }
  long long mid = (L + R) / 2;
  shift(node, L, R);
  updatez(L, mid, 2 * node, l, r, z);
  updatez(mid, R, 2 * node + 1, l, r, z);
  sum[node] = sum[2 * node] + sum[2 * node + 1];
}
void updates(long long L, long long R, long long node, long long l, long long r,
             long double z) {
  if (l <= L && R <= r) {
    shift(node, L, R);
    lazys[node] += z;
    sum[node] += z * (R - L);
    return;
  }
  if (R <= l || r <= L) {
    return;
  }
  long long mid = (L + R) / 2;
  shift(node, L, R);
  updates(L, mid, 2 * node, l, r, z);
  updates(mid, R, 2 * node + 1, l, r, z);
  sum[node] = sum[2 * node] + sum[2 * node + 1];
}
long double finds(long long L, long long R, long long node, long long l,
                  long long r) {
  if (l <= L && R <= r) {
    return sum[node];
  }
  if (R <= l || r <= L) {
    return 0;
  }
  long long mid = (L + R) / 2;
  shift(node, L, R);
  return finds(L, mid, 2 * node, l, r) + finds(mid, R, 2 * node + 1, l, r);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fill(lazyz, lazyz + maxn * 4, 1);
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    long long v;
    cin >> v;
    updates(0, n, 1, i, i + 1, v);
  }
  for (long long i = 0; i < m; i++) {
    long long typ;
    cin >> typ;
    if (typ == 1) {
      long long l1, l2, r1, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      l1--;
      l2--;
      r1--;
      r2--;
      long long f = (r1 - l1 + 1) * (r2 - l2 + 1);
      long double s2 = finds(0, n, 1, l2, r2 + 1) / f;
      long double s1 = finds(0, n, 1, l1, r1 + 1) / f;
      long double p = (r1 - l1) / ((long double)(r1 - l1 + 1));
      updatez(0, n, 1, l1, r1 + 1, p);
      updates(0, n, 1, l1, r1 + 1, s2);
      p = (r2 - l2) / ((long double)(r2 - l2 + 1));
      updatez(0, n, 1, l2, r2 + 1, p);
      updates(0, n, 1, l2, r2 + 1, s1);
    } else {
      long long l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << fixed << setprecision(5) << finds(0, n, 1, l, r + 1) << endl;
    }
  }
}