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 = 100100;
const int MAXSEG = 1 << 18;
int N, Q;
int A[MAXN];
double seg[MAXSEG];
double mult[MAXSEG];
double add[MAXSEG];
void build_tree(int node, int left, int right) {
  if (left == right) {
    seg[node] = A[left];
    return;
  }
  int mid = (left + right) / 2;
  build_tree(2 * node, left, mid);
  build_tree(2 * node + 1, mid + 1, right);
  seg[node] = seg[2 * node] + seg[2 * node + 1];
}
void down(int node, int left, int right) {
  int mid = (left + right) / 2;
  seg[2 * node] = mult[node] * seg[2 * node] + add[node] * (mid - left + 1);
  seg[2 * node + 1] =
      mult[node] * seg[2 * node + 1] + add[node] * (right - mid);
  mult[2 * node] *= mult[node];
  mult[2 * node + 1] *= mult[node];
  add[2 * node] = mult[node] * add[2 * node] + add[node];
  add[2 * node + 1] = mult[node] * add[2 * node + 1] + add[node];
  mult[node] = 1;
  add[node] = 0;
}
void update_mult(int node, int left, int right, int ql, int qr, double val) {
  if (left > right || ql > qr || right < ql || qr < left) return;
  if (ql <= left && right <= qr) {
    seg[node] *= val;
    mult[node] *= val;
    add[node] *= val;
    return;
  }
  down(node, left, right);
  int mid = (left + right) / 2;
  update_mult(2 * node, left, mid, ql, qr, val);
  update_mult(2 * node + 1, mid + 1, right, ql, qr, val);
  seg[node] = seg[2 * node] + seg[2 * node + 1];
}
void update_add(int node, int left, int right, int ql, int qr, double val) {
  if (left > right || ql > qr || right < ql || qr < left) return;
  if (ql <= left && right <= qr) {
    seg[node] += val * (right - left + 1);
    add[node] += val;
    return;
  }
  down(node, left, right);
  int mid = (left + right) / 2;
  update_add(2 * node, left, mid, ql, qr, val);
  update_add(2 * node + 1, mid + 1, right, ql, qr, val);
  seg[node] = seg[2 * node] + seg[2 * node + 1];
}
double query(int node, int left, int right, int ql, int qr) {
  if (left > right || ql > qr || right < ql || qr < left) return 0;
  if (ql <= left && right <= qr) return seg[node];
  down(node, left, right);
  int mid = (left + right) / 2;
  return query(2 * node, left, mid, ql, qr) +
         query(2 * node + 1, mid + 1, right, ql, qr);
}
int main() {
  scanf("%d %d", &N, &Q);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  build_tree(1, 1, N);
  for (int i = 0; i < MAXSEG; i++) mult[i] = 1;
  for (int i = 0, type; i < Q; i++) {
    scanf("%d", &type);
    if (type == 1) {
      int l1, r1, l2, r2;
      scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
      int len1 = r1 - l1 + 1;
      int len2 = r2 - l2 + 1;
      double sum1 = query(1, 1, N, l1, r1);
      double sum2 = query(1, 1, N, l2, r2);
      update_mult(1, 1, N, l1, r1, 1.0 * (len1 - 1) / len1);
      update_mult(1, 1, N, l2, r2, 1.0 * (len2 - 1) / len2);
      update_add(1, 1, N, l1, r1, sum2 / len1 / len2);
      update_add(1, 1, N, l2, r2, sum1 / len1 / len2);
    } else {
      int ql, qr;
      scanf("%d %d", &ql, &qr);
      printf("%.10f\n", query(1, 1, N, ql, qr));
    }
  }
  return 0;
}