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 gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 100000;
const int MAXS = 4 * MAXN;
int n, nq;
int a[MAXN];
double ssum[MAXS], slazymlt[MAXS], slazyadd[MAXS];
void sapply(int x, double mlt, double add, int cnt) {
  ssum[x] = ssum[x] * mlt + add * cnt;
  slazymlt[x] *= mlt;
  slazyadd[x] *= mlt;
  slazyadd[x] += add;
}
void spush(int x, int cnt1, int cnt2) {
  if (slazymlt[x] != 1 || slazyadd[x] != 0)
    sapply(2 * x + 1, slazymlt[x], slazyadd[x], cnt1),
        sapply(2 * x + 2, slazymlt[x], slazyadd[x], cnt2), slazymlt[x] = 1,
                                                           slazyadd[x] = 0;
}
void spull(int x) { ssum[x] = ssum[2 * x + 1] + ssum[2 * x + 2]; }
void sinit(int x, int l, int r) {
  slazymlt[x] = 1, slazyadd[x] = 0;
  if (l == r) {
    ssum[x] = a[l];
  } else {
    int m = l + (r - l) / 2;
    sinit(2 * x + 1, l, m);
    sinit(2 * x + 2, m + 1, r);
    spull(x);
  }
}
double sget(int x, int l, int r, int L, int R) {
  if (L <= l && r <= R) {
    return ssum[x];
  } else {
    int m = l + (r - l) / 2;
    double ret = 0;
    spush(x, m - l + 1, r - m);
    if (L <= m) ret += sget(2 * x + 1, l, m, L, R);
    if (m + 1 <= R) ret += sget(2 * x + 2, m + 1, r, L, R);
    spull(x);
    return ret;
  }
}
void smod(int x, int l, int r, int L, int R, double mlt, double add) {
  if (L <= l && r <= R) {
    sapply(x, mlt, add, r - l + 1);
  } else {
    int m = l + (r - l) / 2;
    spush(x, m - l + 1, r - m);
    if (L <= m) smod(2 * x + 1, l, m, L, R, mlt, add);
    if (m + 1 <= R) smod(2 * x + 2, m + 1, r, L, R, mlt, add);
    spull(x);
  }
}
void run() {
  scanf("%d%d", &n, &nq);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  sinit(0, 0, n - 1);
  for (int qi = (0); qi < (nq); ++qi) {
    int kind;
    scanf("%d", &kind);
    if (kind == 1) {
      int l1, r1, l2, r2;
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      --l1, --r1, --l2, --r2;
      double sum1 = sget(0, 0, n - 1, l1, r1), sum2 = sget(0, 0, n - 1, l2, r2);
      int cnt1 = r1 - l1 + 1, cnt2 = r2 - l2 + 1;
      smod(0, 0, n - 1, l1, r1, 1.0 * (cnt1 - 1) / cnt1, sum2 / cnt1 / cnt2);
      smod(0, 0, n - 1, l2, r2, 1.0 * (cnt2 - 1) / cnt2, sum1 / cnt1 / cnt2);
    }
    if (kind == 2) {
      int l, r;
      scanf("%d%d", &l, &r);
      --l, --r;
      double sum = sget(0, 0, n - 1, l, r);
      printf("%.15lf\n", sum);
    }
  }
}
int main() {
  run();
  return 0;
}