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 maxN = 100005;
int N, Q;
double E[maxN], segTree[maxN * 4];
pair<double, double> lazy[maxN * 4];
void build(int node, int left, int right) {
  lazy[node] = {1, 0};
  if (left == right) {
    segTree[node] = E[left];
    return;
  }
  int mid = (left + right) / 2;
  build(node * 2, left, mid);
  build(node * 2 + 1, mid + 1, right);
  segTree[node] = segTree[node * 2] + segTree[node * 2 + 1];
}
void update_lazy(int node, int left, int right, double& a, double& b) {
  segTree[node] = a * segTree[node] + b * (right - left + 1);
  lazy[node] = {lazy[node].first * a, lazy[node].second * a + b};
}
void update(int node, int left, int right, int& queryL, int& queryR, double& a,
            double& b) {
  if (right < queryL || left > queryR) return;
  if (queryL <= left && right <= queryR) {
    update_lazy(node, left, right, a, b);
    return;
  }
  int mid = (left + right) / 2;
  update_lazy(node * 2, left, mid, lazy[node].first, lazy[node].second);
  update_lazy(node * 2 + 1, mid + 1, right, lazy[node].first,
              lazy[node].second);
  lazy[node] = {1, 0};
  update(node * 2, left, mid, queryL, queryR, a, b);
  update(node * 2 + 1, mid + 1, right, queryL, queryR, a, b);
  segTree[node] = segTree[node * 2] + segTree[node * 2 + 1];
}
double query(int node, int left, int right, int& queryL, int& queryR) {
  if (queryL <= left && right <= queryR) return segTree[node];
  if (left > queryR || right < queryL) return 0;
  int mid = (left + right) / 2;
  update_lazy(node * 2, left, mid, lazy[node].first, lazy[node].second);
  update_lazy(node * 2 + 1, mid + 1, right, lazy[node].first,
              lazy[node].second);
  lazy[node] = {1, 0};
  double lval = query(node * 2, left, mid, queryL, queryR);
  double rval = query(node * 2 + 1, mid + 1, right, queryL, queryR);
  return lval + rval;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> Q;
  for (int i = 1; i <= N; ++i) cin >> E[i];
  build(1, 1, N);
  while (Q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int l1, r1, l2, r2;
      cin >> l1 >> r1 >> l2 >> r2;
      double s1 = query(1, 1, N, l1, r1);
      double s2 = query(1, 1, N, l2, r2);
      s1 /= r1 - l1 + 1;
      s1 /= r2 - l2 + 1;
      s2 /= r1 - l1 + 1;
      s2 /= r2 - l2 + 1;
      double a1 = (double)(r1 - l1) / (r1 - l1 + 1);
      double a2 = (double)(r2 - l2) / (r2 - l2 + 1);
      update(1, 1, N, l1, r1, a1, s2);
      update(1, 1, N, l2, r2, a2, s1);
    } else {
      int l, r;
      cin >> l >> r;
      printf("%.6f\n", query(1, 1, N, l, r));
    }
  }
  return 0;
}