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 = 1e5 + 5;
const double eps = 1e-6;
int a[MAXN];
struct seg {
  struct node {
    double val, mul, add;
  } sum[MAXN << 2];
  int sgn(double x) {
    if (fabs(x) < eps)
      return 0;
    else if (x > eps)
      return 1;
    else if (x < -eps)
      return -1;
  }
  inline void push_up(int rt) {
    sum[rt].val = sum[rt << 1].val + sum[rt << 1 | 1].val;
  }
  inline void push_down(int l, int r, int rt) {
    if (sgn(sum[rt].mul - 1)) {
      sum[rt << 1].val *= sum[rt].mul;
      sum[rt << 1].mul *= sum[rt].mul;
      sum[rt << 1].add *= sum[rt].mul;
      sum[rt << 1 | 1].val *= sum[rt].mul;
      sum[rt << 1 | 1].mul *= sum[rt].mul;
      sum[rt << 1 | 1].add *= sum[rt].mul;
      sum[rt].mul = 1;
    }
    if (sgn(sum[rt].add)) {
      int mid = (l + r) >> 1;
      sum[rt << 1].val += sum[rt].add * (mid - l + 1);
      sum[rt << 1].add += sum[rt].add;
      sum[rt << 1 | 1].val += sum[rt].add * (r - mid);
      sum[rt << 1 | 1].add += sum[rt].add;
      sum[rt].add = 0;
    }
  }
  void build(int l, int r, int rt) {
    sum[rt].mul = 1;
    sum[rt].add = 0;
    if (l == r) {
      sum[rt].val = a[l];
      return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    push_up(rt);
  }
  void update_mul(int L, int R, double val, int l, int r, int rt) {
    if (L <= l && r <= R) {
      sum[rt].mul *= val;
      sum[rt].val *= val;
      sum[rt].add *= val;
      return;
    }
    int mid = (l + r) >> 1;
    push_down(l, r, rt);
    if (L <= mid) {
      update_mul(L, R, val, l, mid, rt << 1);
    }
    if (mid < R) {
      update_mul(L, R, val, mid + 1, r, rt << 1 | 1);
    }
    push_up(rt);
  }
  void update_add(int L, int R, double val, int l, int r, int rt) {
    if (L <= l && r <= R) {
      sum[rt].add += val;
      sum[rt].val += (r - l + 1) * val;
      return;
    }
    int mid = (l + r) >> 1;
    push_down(l, r, rt);
    if (L <= mid) {
      update_add(L, R, val, l, mid, rt << 1);
    }
    if (mid < R) {
      update_add(L, R, val, mid + 1, r, rt << 1 | 1);
    }
    push_up(rt);
  }
  double query(int L, int R, int l, int r, int rt) {
    if (L <= l && r <= R) {
      return sum[rt].val;
    }
    int mid = (l + r) >> 1;
    double ret = 0;
    push_down(l, r, rt);
    if (L <= mid) {
      ret += query(L, R, l, mid, rt << 1);
    }
    if (mid < R) {
      ret += query(L, R, mid + 1, r, rt << 1 | 1);
    }
    return ret;
  }
} se;
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  se.build(1, n, 1);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l1, r1, l2, r2;
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double x1 = se.query(l1, r1, 1, n, 1) / (r1 - l1 + 1) / (r2 - l2 + 1);
      double x2 = se.query(l2, r2, 1, n, 1) / (r2 - l2 + 1) / (r1 - l1 + 1);
      se.update_mul(l1, r1, 1.0 * (r1 - l1) / (r1 - l1 + 1), 1, n, 1);
      se.update_mul(l2, r2, 1.0 * (r2 - l2) / (r2 - l2 + 1), 1, n, 1);
      se.update_add(l1, r1, x2, 1, n, 1);
      se.update_add(l2, r2, x1, 1, n, 1);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%.10lf\n", se.query(l, r, 1, n, 1));
    }
  }
  return 0;
}