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, a[maxn], ty, l1, r1, l2, r2, l, r;
struct node {
  int l, r, mark;
  double val, lzk, lzd;
  inline void update(double k, double d) {
    mark = true;
    val = val * k + (r - l + 1) * d;
    lzk *= k;
    lzd = k * lzd + d;
  }
} seg[maxn << 2];
inline void pushup(int x) {
  seg[x].val = seg[x << 1].val + seg[x << 1 | 1].val;
}
inline void pushdown(int x) {
  bool mm = seg[x].mark;
  if (mm) {
    seg[x].mark = false;
    seg[x << 1].update(seg[x].lzk, seg[x].lzd);
    seg[x << 1 | 1].update(seg[x].lzk, seg[x].lzd);
    seg[x].lzk = 1;
    seg[x].lzd = 0;
  }
}
void build(int x, int l, int r) {
  seg[x].l = l, seg[x].r = r, seg[x].mark = false;
  seg[x].lzk = 1, seg[x].lzd = 0;
  if (l == r) {
    seg[x].val = a[l];
    return;
  }
  int m = (l + r) >> 1;
  build(x << 1, l, m);
  build(x << 1 | 1, m + 1, r);
  pushup(x);
}
void update(int x, int l, int r, double k, double d) {
  int L = seg[x].l, R = seg[x].r;
  if (l <= L && r >= R) {
    seg[x].update(k, d);
    return;
  }
  pushdown(x);
  int m = (L + R) >> 1;
  if (m >= l) update(x << 1, l, r, k, d);
  if (m < r) update(x << 1 | 1, l, r, k, d);
  pushup(x);
}
double query(int x, int l, int r) {
  int L = seg[x].l, R = seg[x].r;
  if (l <= L && r >= R) return seg[x].val;
  int m = (L + R) >> 1;
  double ret = 0;
  pushdown(x);
  if (m >= l) ret += query(x << 1, l, r);
  if (m < r) ret += query(x << 1 | 1, l, r);
  pushup(x);
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  build(1, 1, n);
  while (q--) {
    scanf("%d", &ty);
    if (ty == 1) {
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double v1 = query(1, l1, r1) / (r1 - l1 + 1) / (r2 - l2 + 1);
      double v2 = query(1, l2, r2) / (r2 - l2 + 1) / (r1 - l1 + 1);
      update(1, l1, r1, 1.0 * (r1 - l1) / (r1 - l1 + 1), v2);
      update(1, l2, r2, 1.0 * (r2 - l2) / (r2 - l2 + 1), v1);
    } else {
      scanf("%d%d", &l, &r);
      printf("%.12f\n", query(1, l, r));
    }
  }
  return 0;
}