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 std ::map;
using std ::priority_queue;
using std ::queue;
using std ::set;
using std ::string;
using std ::swap;
using std ::unordered_map;
using std ::vector;
template <typename Tp>
Tp Max(const Tp &a, const Tp &b) {
  return a > b ? a : b;
}
template <typename Tp>
Tp Min(const Tp &a, const Tp &b) {
  return a < b ? a : b;
}
template <typename Tp>
Tp Abs(const Tp &a) {
  return a > 0 ? a : -a;
}
template <typename Tp>
void Read(Tp &x) {
  Tp in = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    in = in * 10 + ch - '0';
    ch = getchar();
  }
  x = in * f;
}
const int SN = 100000 + 10;
double sum[SN << 2], add[SN << 2], mul[SN << 2];
int a[SN];
int n, m;
void pushup(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; }
void pushdown(int l, int r, int rt) {
  int mid = (l + r) >> 1;
  if (mul[rt] != 1) {
    sum[rt << 1] *= mul[rt];
    sum[rt << 1 | 1] *= mul[rt];
    add[rt << 1] *= mul[rt];
    add[rt << 1 | 1] *= mul[rt];
    mul[rt << 1] *= mul[rt];
    mul[rt << 1 | 1] *= mul[rt];
    mul[rt] = 1;
  }
  if (add[rt] != 0) {
    sum[rt << 1] += (mid - l + 1) * add[rt];
    sum[rt << 1 | 1] += (r - mid) * add[rt];
    add[rt << 1] += add[rt];
    add[rt << 1 | 1] += add[rt];
    add[rt] = 0;
  }
}
void build(int l, int r, int rt) {
  mul[rt] = 1;
  if (l == r) {
    sum[rt] = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1), build(mid + 1, r, rt << 1 | 1);
  pushup(rt);
}
void Mul(int QL, int QR, double X, int l, int r, int rt) {
  if (QL <= l && QR >= r) {
    sum[rt] *= X;
    add[rt] *= X;
    mul[rt] *= X;
    return;
  }
  pushdown(l, r, rt);
  int mid = (l + r) >> 1;
  if (QL <= mid) Mul(QL, QR, X, l, mid, rt << 1);
  if (QR > mid) Mul(QL, QR, X, mid + 1, r, rt << 1 | 1);
  pushup(rt);
}
void Add(int QL, int QR, double X, int l, int r, int rt) {
  if (QL <= l && QR >= r) {
    sum[rt] += (r - l + 1) * X;
    add[rt] += X;
    return;
  }
  pushdown(l, r, rt);
  int mid = (l + r) >> 1;
  if (QL <= mid) Add(QL, QR, X, l, mid, rt << 1);
  if (QR > mid) Add(QL, QR, X, mid + 1, r, rt << 1 | 1);
  pushup(rt);
}
double Query(int QL, int QR, int l, int r, int rt) {
  if (QL <= l && QR >= r) return sum[rt];
  pushdown(l, r, rt);
  double ans = 0;
  int mid = (l + r) >> 1;
  if (QL <= mid) ans += Query(QL, QR, l, mid, rt << 1);
  if (QR > mid) ans += Query(QL, QR, mid + 1, r, rt << 1 | 1);
  return ans;
}
int main(int argc, const char *argv[]) {
  Read(n), Read(m);
  for (int i = 1; i <= n; i++) Read(a[i]);
  build(1, n, 1);
  int opt, l1, l2, r1, r2;
  while (m--) {
    Read(opt), Read(l1), Read(r1);
    if (opt == 2)
      printf("%.10lf\n", Query(l1, r1, 1, n, 1));
    else {
      Read(l2), Read(r2);
      long long L = r1 - l1 + 1, R = r2 - l2 + 1;
      double Lsum = Query(l1, r1, 1, n, 1);
      double Rsum = Query(l2, r2, 1, n, 1);
      Mul(l1, r1, 1.0 * (L - 1) / L, 1, n, 1);
      Mul(l2, r2, 1.0 * (R - 1) / R, 1, n, 1);
      Add(l1, r1, Rsum / (1.0 * L * R), 1, n, 1);
      Add(l2, r2, Lsum / (1.0 * L * R), 1, n, 1);
    }
  }
  return 0;
}