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 = 1e5, maxnode = maxn << 2;
int seg[maxn + 5];
double tree[maxnode + 5], cof[maxnode + 5], del[maxnode + 5];
void Build(int x, int l, int r) {
  cof[x] = 1, del[x] = 0;
  if (l == r) {
    tree[x] = seg[l];
    return;
  }
  Build((x << 1), l, ((l + r) >> 1));
  Build((x << 1 | 1), ((l + r) >> 1) + 1, r);
  tree[x] = tree[(x << 1)] + tree[(x << 1 | 1)];
}
void PushDown(int x, int l, int r) {
  cof[(x << 1)] *= cof[x], del[(x << 1)] = del[(x << 1)] * cof[x] + del[x];
  cof[(x << 1 | 1)] *= cof[x],
      del[(x << 1 | 1)] = del[(x << 1 | 1)] * cof[x] + del[x];
  tree[(x << 1)] = tree[(x << 1)] * cof[x] + del[x] * (((l + r) >> 1) - l + 1);
  tree[(x << 1 | 1)] =
      tree[(x << 1 | 1)] * cof[x] + del[x] * (r - ((l + r) >> 1));
  cof[x] = 1, del[x] = 0;
}
void Upd(int x, int l, int r, int ll, int rr, double mu, double ad) {
  if (ll <= l && r <= rr) {
    cof[x] *= mu, del[x] = del[x] * mu + ad;
    tree[x] = tree[x] * mu + ad * (r - l + 1);
  } else {
    PushDown(x, l, r);
    if (ll <= ((l + r) >> 1)) Upd((x << 1), l, ((l + r) >> 1), ll, rr, mu, ad);
    if (((l + r) >> 1) < rr)
      Upd((x << 1 | 1), ((l + r) >> 1) + 1, r, ll, rr, mu, ad);
    tree[x] =
        cof[x] * (tree[(x << 1)] + tree[(x << 1 | 1)]) + del[x] * (r - l + 1);
  }
}
double Query(int x, int l, int r, int ll, int rr) {
  if (ll <= l && r <= rr) return tree[x];
  double ret = 0;
  if (ll <= ((l + r) >> 1)) ret += Query((x << 1), l, ((l + r) >> 1), ll, rr);
  if (((l + r) >> 1) < rr)
    ret += Query((x << 1 | 1), ((l + r) >> 1) + 1, r, ll, rr);
  return cof[x] * ret + del[x] * (min(rr, r) - max(ll, l) + 1);
}
int main() {
  int n, T, od, l1, r1, l2, r2;
  scanf("%d%d", &n, &T);
  for (int i = 1; i <= n; ++i) scanf("%d", seg + i);
  Build(1, 1, n);
  while (T--) {
    scanf("%d", &od);
    if (od == 1) {
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      int len1 = (r1 - l1 + 1), len2 = (r2 - l2 + 1);
      double sum1 = Query(1, 1, n, l1, r1);
      double sum2 = Query(1, 1, n, l2, r2);
      Upd(1, 1, n, l1, r1, double(len1 - 1) / len1, sum2 / len1 / len2);
      Upd(1, 1, n, l2, r2, double(len2 - 1) / len2, sum1 / len1 / len2);
    } else {
      scanf("%d%d", &l1, &r1);
      printf("%.7f\n", Query(1, 1, n, l1, r1));
    }
  }
  return 0;
}