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 N = 1e5 + 5;
int n, q, fa[N];
double a[N];
struct Tree {
  int l, r, lz;
  double val, mul, add;
} tr[N * 4];
void build(int l, int r, int t) {
  tr[t].l = l;
  tr[t].r = r;
  tr[t].lz = tr[t].val = tr[t].add = 0;
  tr[t].mul = 1;
  if (l == r) {
    fa[l] = t;
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, t << 1);
  build(mid + 1, r, (t << 1) | 1);
}
void pushDown(int t) {
  if (tr[t].lz) {
    tr[t << 1].val = tr[t].add * (tr[t << 1].r - tr[t << 1].l + 1) +
                     tr[t << 1].val * tr[t].mul;
    tr[t << 1].mul *= tr[t].mul;
    tr[t << 1].add *= tr[t].mul;
    tr[t << 1].add += tr[t].add;
    tr[(t << 1) | 1].val =
        tr[t].add * (tr[(t << 1) | 1].r - tr[(t << 1) | 1].l + 1) +
        tr[(t << 1) | 1].val * tr[t].mul;
    tr[(t << 1) | 1].mul *= tr[t].mul;
    tr[(t << 1) | 1].add *= tr[t].mul;
    tr[(t << 1) | 1].add += tr[t].add;
    tr[t << 1].lz = tr[(t << 1) | 1].lz = 1;
    tr[t].lz = tr[t].add = 0;
    tr[t].mul = 1;
  }
}
void pushUp(int t) { tr[t].val = tr[t << 1].val + tr[(t << 1) | 1].val; }
void update(int l, int r, int t, double mul, double add) {
  if (l <= tr[t].l && tr[t].r <= r) {
    tr[t].val = add * (tr[t].r - tr[t].l + 1) + tr[t].val * mul;
    tr[t].lz = 1;
    tr[t].mul *= mul;
    tr[t].add *= mul;
    tr[t].add += add;
    return;
  }
  pushDown(t);
  int mid = (tr[t].l + tr[t].r) / 2;
  if (l <= mid) update(l, r, t << 1, mul, add);
  if (r > mid) update(l, r, (t << 1) | 1, mul, add);
  pushUp(t);
}
double query(int l, int r, int t) {
  if (l <= tr[t].l && tr[t].r <= r) return tr[t].val;
  pushDown(t);
  double ret = 0;
  int mid = (tr[t].l + tr[t].r) / 2;
  if (l <= mid) ret += query(l, r, t << 1);
  if (r > mid) ret += query(l, r, (t << 1) | 1);
  return ret;
}
void solve() {
  scanf("%d%d", &n, &q);
  build(1, n, 1);
  for (int i = 1; i <= n; i++) {
    scanf("%lf", &a[i]);
    update(i, i, 1, 1, a[i]);
  }
  while (q--) {
    int ty;
    scanf("%d", &ty);
    if (ty == 1) {
      int l1, r1, l2, r2;
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double avg1 = query(l1, r1, 1) / double(r1 - l1 + 1);
      double avg2 = query(l2, r2, 1) / double(r2 - l2 + 1);
      update(l1, r1, 1, double(r1 - l1) / double(r1 - l1 + 1),
             avg2 / double(r1 - l1 + 1));
      update(l2, r2, 1, double(r2 - l2) / double(r2 - l2 + 1),
             avg1 / double(r2 - l2 + 1));
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%.7f\n", query(l, r, 1));
    }
  }
}
int main() {
  solve();
  return 0;
}