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;
struct node {
  double m, c, sum, add;
  bool changed;
} st[(1 << 18)];
double a[100010];
void construct(int L, int R, int curr) {
  st[curr].m = 1;
  st[curr].c = 0;
  st[curr].add = 0;
  st[curr].changed = false;
  if (L == R) {
    st[curr].sum = a[L];
    return;
  }
  int mid = ((L + R) >> 1), Lsn = (curr << 1), Rsn = (Lsn | 1);
  construct(L, mid, Lsn);
  construct(mid + 1, R, Rsn);
  st[curr].sum = st[Lsn].sum + st[Rsn].sum;
}
void updateNode(int curr, double factor1, double factor2, double len) {
  st[curr].m *= factor1;
  st[curr].c *= factor1;
  st[curr].c += factor2;
  st[curr].sum *= factor1;
  st[curr].sum += factor2 * len;
}
void update(int i, int j, int L, int R, int curr, double sum, double len1,
            double len2) {
  if (i == L && j == R) {
    double factor1 = (1 - 1.0 / len1), factor2 = sum / (len1 * len2);
    updateNode(curr, factor1, factor2, R - L + 1);
    return;
  }
  int mid = ((L + R) >> 1), Lsn = (curr << 1), Rsn = (Lsn | 1);
  if (st[curr].m != 1 || st[curr].c != 0) {
    updateNode(Lsn, st[curr].m, st[curr].c, mid - L + 1);
    updateNode(Rsn, st[curr].m, st[curr].c, R - mid);
    st[curr].m = 1;
    st[curr].c = 0;
  }
  if (i > mid)
    update(i, j, mid + 1, R, Rsn, sum, len1, len2);
  else if (j <= mid)
    update(i, j, L, mid, Lsn, sum, len1, len2);
  else {
    update(i, mid, L, mid, Lsn, sum, len1, len2);
    update(mid + 1, j, mid + 1, R, Rsn, sum, len1, len2);
  }
  st[curr].sum = st[Lsn].sum + st[Rsn].sum;
}
double getSum(int i, int j, int L, int R, int curr) {
  if (i == L && j == R) {
    return st[curr].sum;
  }
  int mid = ((L + R) >> 1), Lsn = (curr << 1), Rsn = (Lsn | 1);
  if (st[curr].m != 1 || st[curr].c != 0) {
    updateNode(Lsn, st[curr].m, st[curr].c, mid - L + 1);
    updateNode(Rsn, st[curr].m, st[curr].c, R - mid);
    st[curr].m = 1;
    st[curr].c = 0;
  }
  if (i > mid) return getSum(i, j, mid + 1, R, Rsn);
  if (j <= mid) return getSum(i, j, L, mid, Lsn);
  return getSum(i, mid, L, mid, Lsn) + getSum(mid + 1, j, mid + 1, R, Rsn);
}
void action1(int l1, int r1, int l2, int r2, int n) {
  double sum1 = getSum(l1, r1, 1, n, 1), sum2 = getSum(l2, r2, 1, n, 1);
  double len1 = r1 - l1 + 1, len2 = r2 - l2 + 1;
  update(l1, r1, 1, n, 1, sum2, len1, len2);
  update(l2, r2, 1, n, 1, sum1, len2, len1);
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%lf", &a[i]);
  }
  construct(1, n, 1);
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int l1, l2, r1, r2;
      scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
      action1(l1, r1, l2, r2, n);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      printf("%0.6lf\n", getSum(l, r, 1, n, 1));
    }
  }
  return 0;
}