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;
inline void read(int &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
inline void read(long long &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
int n, m;
int len[120000 * 7];
double tree[120000 * 7];
double add[120000 * 7], mul[120000 * 7];
int a[120000];
void update(int p) { tree[p] = tree[p * 2] + tree[p * 2 + 1]; }
void build(int p, int l, int r) {
  mul[p] = 1;
  add[p] = 0;
  len[p] = r - l + 1;
  if (l == r) {
    tree[p] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build(p * 2, l, mid);
  build(p * 2 + 1, mid + 1, r);
  update(p);
}
void pushdown(int p) {
  tree[p * 2] = tree[p * 2] * mul[p] + add[p] * len[p * 2];
  add[p * 2] = add[p * 2] * mul[p] + add[p];
  mul[p * 2] = mul[p * 2] * mul[p];
  tree[p * 2 + 1] = tree[p * 2 + 1] * mul[p] + add[p] * len[p * 2 + 1];
  add[p * 2 + 1] = add[p * 2 + 1] * mul[p] + add[p];
  mul[p * 2 + 1] = mul[p * 2 + 1] * mul[p];
  mul[p] = 1;
  add[p] = 0;
}
void change(int p, int l, int r, int a, int b, double mull, double addd) {
  pushdown(p);
  if ((l == a) && (r == b)) {
    tree[p] = tree[p] * mull + addd * len[p];
    add[p] = add[p] * mull + addd;
    mul[p] = mul[p] * mull;
    return;
  }
  int mid = (l + r) / 2;
  if (b <= mid)
    change(p * 2, l, mid, a, b, mull, addd);
  else if (a > mid)
    change(p * 2 + 1, mid + 1, r, a, b, mull, addd);
  else {
    change(p * 2, l, mid, a, mid, mull, addd);
    change(p * 2 + 1, mid + 1, r, mid + 1, b, mull, addd);
  }
  update(p);
}
double query(int p, int l, int r, int a, int b) {
  if ((l == a) && (r == b)) return tree[p];
  pushdown(p);
  int mid = (l + r) / 2;
  if (b <= mid)
    return query(p * 2, l, mid, a, b);
  else if (a > mid)
    return query(p * 2 + 1, mid + 1, r, a, b);
  else
    return query(p * 2, l, mid, a, mid) +
           query(p * 2 + 1, mid + 1, r, mid + 1, b);
}
int main() {
  read(n);
  read(m);
  for (int i = 1; i <= n; i++) read(a[i]);
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int op;
    read(op);
    if (op == 1) {
      int l1, r1, l2, r2;
      read(l1);
      read(r1);
      read(l2);
      read(r2);
      double t1 = query(1, 1, n, l1, r1);
      double t2 = query(1, 1, n, l2, r2);
      change(1, 1, n, l1, r1, 1.0 * (r1 - l1) / (r1 - l1 + 1),
             t2 / (r2 - l2 + 1) / (r1 - l1 + 1));
      change(1, 1, n, l2, r2, 1.0 * (r2 - l2) / (r2 - l2 + 1),
             t1 / (r1 - l1 + 1) / (r2 - l2 + 1));
    } else {
      int l, r;
      read(l);
      read(r);
      printf("%.6f\n", query(1, 1, n, l, r));
    }
  }
}