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 = 410000;
struct node {
  double lm, la, sum;
  int l, r;
} tr[maxn];
int n, q;
int a[maxn];
void update(int k) { tr[k].sum = tr[k << 1].sum + tr[k << 1 | 1].sum; }
void pushdown1(int k) {
  tr[k << 1].sum += tr[k].la * (tr[k << 1].r - tr[k << 1].l + 1),
      tr[k << 1].la += tr[k].la;
  tr[k << 1 | 1].sum += tr[k].la * (tr[k << 1 | 1].r - tr[k << 1 | 1].l + 1),
      tr[k << 1 | 1].la += tr[k].la;
  tr[k].la = 0;
}
void pushdown2(int k) {
  tr[k << 1].sum *= tr[k].lm, tr[k << 1].la *= tr[k].lm,
      tr[k << 1].lm *= tr[k].lm;
  tr[k << 1 | 1].sum *= tr[k].lm, tr[k << 1 | 1].la *= tr[k].lm,
      tr[k << 1 | 1].lm *= tr[k].lm;
  tr[k].lm = 1;
}
void build(int k, int l, int r) {
  tr[k].lm = 1, tr[k].la = 0;
  tr[k].l = l, tr[k].r = r;
  if (l == r) {
    tr[k].sum = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
  update(k);
}
double ask(int k, int l, int r, int a, int b) {
  if (l == a && r == b) return tr[k].sum;
  int mid = (l + r) / 2;
  pushdown2(k);
  pushdown1(k);
  double res = 0;
  if (b <= mid)
    res = ask(k << 1, l, mid, a, b);
  else if (a > mid)
    res = ask(k << 1 | 1, mid + 1, r, a, b);
  else
    res = ask(k << 1, l, mid, a, mid) + ask(k << 1 | 1, mid + 1, r, mid + 1, b);
  update(k);
  return res;
}
void change1(int k, int l, int r, int a, int b, double c) {
  if (l == a && r == b) {
    tr[k].sum += c * (double)(r - l + 1);
    tr[k].la += c;
    return;
  }
  int mid = (l + r) / 2;
  pushdown2(k);
  pushdown1(k);
  if (b <= mid)
    change1(k << 1, l, mid, a, b, c);
  else if (a > mid)
    change1(k << 1 | 1, mid + 1, r, a, b, c);
  else
    change1(k << 1, l, mid, a, mid, c),
        change1(k << 1 | 1, mid + 1, r, mid + 1, b, c);
  update(k);
}
void change2(int k, int l, int r, int a, int b, double c) {
  if (l == a && r == b) {
    tr[k].sum *= c;
    tr[k].lm *= c;
    tr[k].la *= c;
    return;
  }
  int mid = (l + r) / 2;
  pushdown2(k);
  pushdown1(k);
  if (b <= mid)
    change2(k << 1, l, mid, a, b, c);
  else if (a > mid)
    change2(k << 1 | 1, mid + 1, r, a, b, c);
  else
    change2(k << 1, l, mid, a, mid, c),
        change2(k << 1 | 1, mid + 1, r, mid + 1, b, c);
  update(k);
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build(1, 1, n);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l1, r1, l2, r2;
      scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
      int len1 = r1 - l1 + 1, len2 = r2 - l2 + 1;
      double sum1 = ask(1, 1, n, l1, r1);
      double sum2 = ask(1, 1, n, l2, r2);
      change2(1, 1, n, l1, r1, (double)(len1 - 1) / (double)len1);
      change2(1, 1, n, l2, r2, (double)(len2 - 1) / (double)len2);
      change1(1, 1, n, l1, r1, (double)sum2 / (double)len2 / (double)len1);
      change1(1, 1, n, l2, r2, (double)sum1 / (double)len1 / (double)len2);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      printf("%.8f\n", ask(1, 1, n, l, r));
    }
  }
  return 0;
}