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 double EPS = 1e-8;
double arr[100003];
double segtree[400003];
double lazy1[400003], lazy2[400003];
void ceklazy(int pos, int kir, int kan) {
  double sz = (kan - kir + 1);
  if (lazy1[pos] > EPS) {
    double siz = lazy2[pos];
    double val = lazy1[pos];
    segtree[pos] = segtree[pos] * siz + sz * val;
    if (kir != kan) {
      double mid = (kir + kan) / 2;
      lazy2[pos * 2] *= siz;
      lazy2[pos * 2 + 1] *= siz;
      lazy1[pos * 2] = lazy1[pos * 2] * siz + val;
      lazy1[pos * 2 + 1] = lazy1[pos * 2 + 1] * siz + val;
    }
    lazy1[pos] = 0;
    lazy2[pos] = 1;
  }
}
void build(int pos, int kir, int kan) {
  if (kir == kan)
    segtree[pos] = arr[kir];
  else {
    int mid = (kir + kan) / 2;
    build(pos * 2, kir, mid);
    build(pos * 2 + 1, mid + 1, kan);
    segtree[pos] = segtree[pos * 2] + segtree[pos * 2 + 1];
  }
}
void update(int pos, int kir, int kan, int qkir, int qkan, double val,
            double siz) {
  double sz = (kan - kir + 1);
  ceklazy(pos, kir, kan);
  if (kir > qkan || qkir > kan)
    return;
  else if (qkir <= kir && kan <= qkan) {
    segtree[pos] += (sz / siz) * (val - segtree[pos] / sz);
    if (kir != kan) {
      double mid = (kir + kan) / 2;
      lazy2[pos * 2] *= (siz - 1) / siz;
      lazy2[pos * 2 + 1] *= (siz - 1) / siz;
      lazy1[pos * 2] = lazy1[pos * 2] * (siz - 1) / siz + (val / siz);
      lazy1[pos * 2 + 1] = lazy1[pos * 2 + 1] * (siz - 1) / siz + (val / siz);
    }
  } else {
    int mid = (kir + kan) / 2;
    update(pos * 2, kir, mid, qkir, qkan, val, siz);
    update(pos * 2 + 1, mid + 1, kan, qkir, qkan, val, siz);
    segtree[pos] = segtree[pos * 2] + segtree[pos * 2 + 1];
  }
}
double query(int pos, int kir, int kan, int qkir, int qkan) {
  ceklazy(pos, kir, kan);
  if (kir > qkan || qkir > kan)
    return 0;
  else if (qkir <= kir && kan <= qkan)
    return segtree[pos];
  else {
    int mid = (kir + kan) / 2;
    return query(pos * 2, kir, mid, qkir, qkan) +
           query(pos * 2 + 1, mid + 1, kan, qkir, qkan);
  }
}
void print(int n) {
  printf("SEGTREE : ");
  for (int i = 1; i <= 2 * n; i++) printf("%.3lf ", segtree[i]);
  printf("\n");
  printf("LAZY1 	: ");
  for (int i = 1; i <= 2 * n; i++) printf("%.3lf ", lazy1[i]);
  printf("\n");
  printf("LAZY2 	: ");
  for (int i = 1; i <= 2 * n; i++) printf("%.3lf ", lazy2[i]);
  printf("\n");
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%lf", &arr[i]);
  for (int i = 1; i <= 4 * n; i++) lazy2[i] = 1;
  build(1, 1, n);
  while (q--) {
    int task;
    scanf("%d", &task);
    if (task == 1) {
      double x1, x2, y1, y2;
      scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);
      double q1 = query(1, 1, n, x1, y1);
      double q2 = query(1, 1, n, x2, y2);
      double sz1 = (y1 - x1 + 1);
      double sz2 = (y2 - x2 + 1);
      update(1, 1, n, x1, y1, q2 / sz2, sz1);
      update(1, 1, n, x2, y2, q1 / sz1, sz2);
    } else {
      double x, y;
      scanf("%lf %lf", &x, &y);
      printf("%.6lf\n", query(1, 1, n, x, y));
    }
  }
}