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;
long long mod = 1000000007LL;
long long large = 1000000000000000000LL;
vector<long double> addv, sum, multi;
int ul, ur;
long double uv, um;
void pushdown(int o) {
  multi[o * 2] *= multi[o];
  multi[o * 2 + 1] *= multi[o];
  addv[o * 2] *= multi[o];
  addv[o * 2 + 1] *= multi[o];
  multi[o] = 1;
  addv[o * 2] += addv[o];
  addv[o * 2 + 1] += addv[o];
  addv[o] = 0;
}
void maintain(int o, int l, int r) {
  if (l == r) {
    sum[o] = addv[o];
  } else {
    sum[o] = multi[o] * (sum[o * 2] + sum[o * 2 + 1]) + addv[o] * (r - l + 1);
  }
}
void update(int o, int l, int r) {
  if (ul <= l && ur >= r) {
    addv[o] *= um;
    multi[o] *= um;
    addv[o] += uv;
  } else {
    pushdown(o);
    int m = (l + r) / 2;
    if (ul <= m)
      update(o * 2, l, m);
    else
      maintain(o * 2, l, m);
    if (ur > m)
      update(o * 2 + 1, m + 1, r);
    else
      maintain(o * 2 + 1, m + 1, r);
  }
  maintain(o, l, r);
}
int ql, qr;
long double re;
void query(int o, int l, int r, long double add, long double mul) {
  if (ql <= l && qr >= r) {
    re += sum[o] * mul + add * (r - l + 1);
  } else {
    int m = (l + r) / 2;
    if (ql <= m) query(o * 2, l, m, add + addv[o] * mul, mul * multi[o]);
    if (qr > m) query(o * 2 + 1, m + 1, r, add + addv[o] * mul, mul * multi[o]);
  }
}
int main() {
  int n, q;
  cin >> n >> q;
  sum.assign(4 * n + 10, 0);
  addv = sum;
  multi.assign(4 * n + 10, 1);
  for (int i = 0; i < n; i++) {
    ul = ur = i;
    cin >> uv;
    um = 1;
    update(1, 0, n - 1);
  }
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int l1, r1, l2, r2;
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      l1--;
      r1--;
      l2--;
      r2--;
      ql = l1;
      qr = r1;
      re = 0;
      query(1, 0, n - 1, 0, 1);
      long double v1 = re;
      v1 /= (r1 - l1 + 1);
      v1 /= (r2 - l2 + 1);
      ql = l2;
      qr = r2;
      re = 0;
      query(1, 0, n - 1, 0, 1);
      long double v2 = re;
      v2 /= (r2 - l2 + 1);
      v2 /= (r1 - l1 + 1);
      ul = l1;
      ur = r1;
      uv = v2;
      um = (long double)(r1 - l1) / (r1 - l1 + 1);
      update(1, 0, n - 1);
      ul = l2;
      ur = r2;
      uv = v1;
      um = (long double)(r2 - l2) / (r2 - l2 + 1);
      update(1, 0, n - 1);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      l--;
      r--;
      ql = l;
      qr = r;
      re = 0;
      query(1, 0, n - 1, 0, 1);
      cout << setprecision(8) << fixed << re << endl;
    }
  }
  return 0;
}