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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 100005;
int n;
double t[N * 4];
double lazg[N * 4], lazb[N * 4];
void shi(int tl, int tr, int pos) {
  int m = (tl + tr) / 2;
  t[pos * 2] = t[pos * 2] * lazb[pos] + (m - tl + 1) * lazg[pos];
  lazb[pos * 2] *= lazb[pos];
  lazg[pos * 2] = lazg[pos * 2] * lazb[pos] + lazg[pos];
  t[pos * 2 + 1] = t[pos * 2 + 1] * lazb[pos] + (tr - m) * lazg[pos];
  lazb[pos * 2 + 1] *= lazb[pos];
  lazg[pos * 2 + 1] = lazg[pos * 2 + 1] * lazb[pos] + lazg[pos];
  lazb[pos] = 1;
  lazg[pos] = 0;
}
void ubd(int tl, int tr, int l, int r, double x, double y, int pos) {
  if (l > r) return;
  if (tl == l && tr == r) {
    t[pos] = t[pos] * x + (r - l + 1) * y;
    lazb[pos] *= x;
    lazg[pos] = lazg[pos] * x + y;
    return;
  }
  shi(tl, tr, pos);
  int m = (tl + tr) / 2;
  ubd(tl, m, l, min(m, r), x, y, pos * 2);
  ubd(m + 1, tr, max(m + 1, l), r, x, y, pos * 2 + 1);
  t[pos] = t[pos * 2] + t[pos * 2 + 1];
}
double qry(int tl, int tr, int l, int r, int pos) {
  if (l > r) return 0;
  if (tl == l && tr == r) return t[pos];
  shi(tl, tr, pos);
  int m = (tl + tr) / 2;
  return (qry(tl, m, l, min(m, r), pos * 2) +
          qry(m + 1, tr, max(m + 1, l), r, pos * 2 + 1));
}
void solv() {
  for (int i = 0; i < N * 4; ++i) lazb[i] = 1;
  int qq;
  scanf("%d%d", &n, &qq);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    ubd(1, n, i, i, 1, x, 1);
  }
  while (qq--) {
    int ty;
    scanf("%d", &ty);
    if (ty == 1) {
      int l1, r1, l2, r2;
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double d1 = r1 - l1 + 1;
      double d2 = r2 - l2 + 1;
      double s1 = qry(1, n, l1, r1, 1);
      double s2 = qry(1, n, l2, r2, 1);
      ubd(1, n, l1, r1, (d1 - 1) / d1, (s2 / d2) / d1, 1);
      ubd(1, n, l2, r2, (d2 - 1) / d2, (s1 / d1) / d2, 1);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%.9f\n", qry(1, n, l, r, 1));
    }
  }
}
int main() {
  solv();
  return 0;
}