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;
int n, q;
const int cc = 1 << 17;
int tab[cc];
struct node {
  double sum = 0;
  double mod[2] = {0.0, 1.0};
  int sizee = 0;
};
struct Tree {
  int rg = cc;
  node nodes[2 * cc];
  double build(int i, int start, int meta) {
    nodes[i].sizee = meta - start + 1;
    if (start == meta) {
      nodes[i].sum = tab[i - rg];
      return nodes[i].sum;
    }
    int s = start + meta;
    s /= 2;
    nodes[i].sum = build(i * 2, start, s) + build(i * 2 + 1, s + 1, meta);
    return nodes[i].sum;
  }
  void inline lazy_sum(int i, double val) {
    if (nodes[i].sizee == 1) {
      return;
    }
    int ind[] = {i * 2, i * 2 + 1};
    for (auto &x : ind) {
      nodes[x].sum += (nodes[x].sizee) * val;
    }
  }
  void inline lazy_il(int i, double val) {
    if (nodes[i].sizee == 1) {
      return;
    }
    int ind[] = {i * 2, i * 2 + 1};
    for (auto &x : ind) {
      nodes[x].sum *= val;
    }
  }
  void inline lazy(int i) {
    for (int j = 1; j >= 0; --j) {
      if (nodes[i].mod[j] == j) continue;
      int ind[] = {i * 2, i * 2 + 1};
      if (nodes[i].sizee > 2) {
        for (auto &x : ind) {
          if (j == 0) {
            nodes[x].mod[j] += nodes[i].mod[j];
          } else {
            nodes[x].mod[1] *= nodes[i].mod[j];
            nodes[x].mod[0] *= nodes[i].mod[j];
          }
        }
      }
      if (j == 0) {
        lazy_sum(i, nodes[i].mod[j]);
      } else {
        lazy_il(i, nodes[i].mod[j]);
      }
      nodes[i].mod[j] = j;
    }
  }
  double add(int i, int a, int b, int start, int meta, double val) {
    lazy(i);
    if (a <= start and b >= meta) {
      nodes[i].sum += val * (nodes[i].sizee);
      if (nodes[i].sizee != 1) nodes[i].mod[0] = val;
      return nodes[i].sum;
    }
    if (b < start or a > meta) {
      return nodes[i].sum;
    }
    int s = start + meta;
    s /= 2;
    nodes[i].sum = add(i * 2, a, b, start, s, val) +
                   add(i * 2 + 1, a, b, s + 1, meta, val);
    return nodes[i].sum;
  }
  double mul(int i, int a, int b, int start, int meta, double val) {
    lazy(i);
    if (a <= start and b >= meta) {
      nodes[i].sum *= val;
      if (nodes[i].sizee != 1) nodes[i].mod[1] = val;
      return nodes[i].sum;
    }
    if (b < start or a > meta) {
      return nodes[i].sum;
    }
    int s = start + meta;
    s /= 2;
    nodes[i].sum = mul(i * 2, a, b, start, s, val) +
                   mul(i * 2 + 1, a, b, s + 1, meta, val);
    return nodes[i].sum;
  }
  double query(int i, int a, int b, int start, int meta) {
    lazy(i);
    if (a <= start and b >= meta) {
      return nodes[i].sum;
    }
    if (b < start or a > meta) {
      return 0;
    }
    int s = start + meta;
    s /= 2;
    double l = query(i * 2, a, b, start, s);
    double r = query(i * 2 + 1, a, b, s + 1, meta);
    return l + r;
  }
};
Tree tree;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 0; i < cc; ++i) tab[i] = 0;
  for (int i = 0; i < n; ++i) cin >> tab[i];
  tree.build(1, 1, cc);
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int a, b, aa, bb;
      cin >> a >> b >> aa >> bb;
      double left = tree.query(1, a, b, 1, cc);
      double right = tree.query(1, aa, bb, 1, cc);
      tree.mul(1, a, b, 1, cc, 1.0 - 1.0 / (b - a + 1));
      tree.mul(1, aa, bb, 1, cc, 1.0 - 1.0 / (bb - aa + 1));
      double wsp = (1.0 / (b - a + 1)) * (1.0 / (bb - aa + 1));
      tree.add(1, a, b, 1, cc, wsp * right);
      tree.add(1, aa, bb, 1, cc, wsp * left);
    } else {
      int a, b;
      cin >> a >> b;
      cout << setprecision(5) << tree.query(1, a, b, 1, cc) << "\n";
    }
  }
}