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;
long long modpow(long long base, long long exponent, long long modulus) {
  if (base == 0 && exponent == 0) return 0;
  long long result = 1;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = (result * base) % modulus;
    exponent = exponent >> 1;
    base = (base * base) % modulus;
  }
  return result;
}
int arr[200110];
struct nodes {
  long double sum, lazymul, lazysum;
  int kids, lazy;
  void split(nodes& left, nodes& right) {
    if (lazy) {
      left.sum *= lazymul;
      left.sum += (lazysum * left.kids);
      right.sum *= lazymul;
      right.sum += (lazysum * right.kids);
      left.lazymul *= lazymul;
      left.lazysum *= lazymul;
      left.lazysum += lazysum;
      right.lazymul *= lazymul;
      right.lazysum *= lazymul;
      right.lazysum += lazysum;
      right.lazy = left.lazy = 1;
    }
    lazy = 0;
    lazysum = 0;
    lazymul = 1;
  }
};
struct SegmentTree {
  vector<nodes> tree;
  nodes identity;
  SegmentTree(void) {
    identity.sum = 0;
    identity.lazymul = 1;
    identity.lazysum = 0;
    identity.lazy = identity.kids = 0;
    tree.resize(200110 * 4);
  }
  nodes merge(nodes& l, nodes& r) {
    nodes ret;
    ret.sum = l.sum + r.sum;
    ret.lazysum = 0;
    ret.lazymul = 1;
    ret.lazy = 0;
    ret.kids = l.kids + r.kids;
    return ret;
  }
  void build(int node, int start, int end) {
    if (start == end) {
      tree[node].sum = arr[start];
      tree[node].lazysum = 0;
      tree[node].lazymul = 1;
      tree[node].kids = 1;
      tree[node].lazy = 0;
      return;
    }
    int mid = (start + end) / 2;
    build((node << 1), start, mid);
    build(((node << 1) + 1), mid + 1, end);
    tree[node] = merge(tree[(node << 1)], tree[((node << 1) + 1)]);
  }
  void update(int node, int start, int end, int left, int right,
              long double mul, long double sum) {
    if (start > end or left > end or right < start) return;
    tree[node].split(tree[(node << 1)], tree[((node << 1) + 1)]);
    if (start >= left and end <= right) {
      tree[node].sum *= mul;
      tree[node].sum += (sum * tree[node].kids);
      tree[node].lazymul = mul;
      tree[node].lazysum = sum;
      tree[node].lazy = 1;
      return;
    }
    int mid = (start + end) / 2;
    update((node << 1), start, mid, left, right, mul, sum);
    update(((node << 1) + 1), mid + 1, end, left, right, mul, sum);
    tree[node] = merge(tree[(node << 1)], tree[((node << 1) + 1)]);
  }
  nodes query(int node, int start, int end, int left, int right) {
    if (start > end or left > end or right < start) return identity;
    tree[node].split(tree[(node << 1)], tree[((node << 1) + 1)]);
    if (start >= left and end <= right) return tree[node];
    int mid = (start + end) / 2;
    nodes a, b, ret;
    a = query((node << 1), start, mid, left, right);
    b = query(((node << 1) + 1), mid + 1, end, left, right);
    ret = merge(a, b);
    return ret;
  }
};
SegmentTree P;
int N, Q;
inline void ReadInput(void) {
  scanf("%d", &N);
  scanf("%d", &Q);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &arr[i]);
  }
}
inline void solve(void) {
  P.build(1, 1, N);
  while (Q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l1, r1, l2, r2;
      scanf("%d", &l1);
      scanf("%d", &r1);
      scanf("%d", &l2);
      scanf("%d", &r2);
      long double lsum = P.query(1, 1, N, l1, r1).sum;
      long double rsum = P.query(1, 1, N, l2, r2).sum;
      long double mul = (double)(r1 - l1) / (double)(r1 - l1 + 1);
      long double sum = (rsum / (r2 - l2 + 1)) / (double)(r1 - l1 + 1);
      P.update(1, 1, N, l1, r1, mul, sum);
      mul = (double)(r2 - l2) / (double)(r2 - l2 + 1);
      sum = (lsum / (r1 - l1 + 1)) / (double)(r2 - l2 + 1);
      P.update(1, 1, N, l2, r2, mul, sum);
    } else {
      int l, r;
      scanf("%d", &l);
      scanf("%d", &r);
      double ans = P.query(1, 1, N, l, r).sum;
      printf("%.9lf\n", ans);
    }
  }
}
inline void Refresh(void) {}
int main() {
  ReadInput();
  solve();
  return 0;
}