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 = 200005;
double sum[maxn << 2];
double pls[maxn << 2];
double mul[maxn << 2];
int n, m;
inline void pushup(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; }
inline void build(int rt, int l, int r) {
  if (l == r) {
    scanf("%lf", &sum[rt]);
    return;
  }
  mul[rt] = 1;
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  pushup(rt);
}
inline void pushdown(int rt, int l, int r) {
  int mid = (l + r) >> 1;
  sum[rt << 1] *= mul[rt];
  mul[rt << 1] *= mul[rt];
  pls[rt << 1] *= mul[rt];
  sum[rt << 1 | 1] *= mul[rt];
  mul[rt << 1 | 1] *= mul[rt];
  pls[rt << 1 | 1] *= mul[rt];
  sum[rt << 1] += pls[rt] * (mid - l + 1);
  pls[rt << 1] += pls[rt];
  sum[rt << 1 | 1] += pls[rt] * (r - mid);
  pls[rt << 1 | 1] += pls[rt];
  pls[rt] = 0;
  mul[rt] = 1;
}
inline void add_qj(int rt, int l, int r, int ql, int qr, double k) {
  if (ql <= l && r <= qr) {
    sum[rt] += k * (r - l + 1);
    pls[rt] += k;
    return;
  }
  pushdown(rt, l, r);
  int mid = (l + r) >> 1;
  if (ql <= mid) add_qj(rt << 1, l, mid, ql, qr, k);
  if (qr > mid) add_qj(rt << 1 | 1, mid + 1, r, ql, qr, k);
  pushup(rt);
}
inline void mul_qj(int rt, int l, int r, int ql, int qr, double k) {
  if (ql <= l && r <= qr) {
    sum[rt] *= k;
    mul[rt] *= k;
    pls[rt] *= k;
    return;
  }
  pushdown(rt, l, r);
  int mid = (l + r) >> 1;
  if (ql <= mid) mul_qj(rt << 1, l, mid, ql, qr, k);
  if (qr > mid) mul_qj(rt << 1 | 1, mid + 1, r, ql, qr, k);
  pushup(rt);
}
double query(int rt, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return sum[rt];
  pushdown(rt, l, r);
  int mid = (l + r) >> 1;
  double sum1 = 0, sum2 = 0;
  if (ql <= mid) sum1 += query(rt << 1, l, mid, ql, qr);
  if (qr > mid) sum2 += query(rt << 1 | 1, mid + 1, r, ql, qr);
  return sum1 + sum2;
}
int main() {
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  while (m--) {
    int opt;
    scanf("%d", &opt);
    if (opt == 1) {
      int a, b, c, d;
      scanf("%d%d%d%d", &a, &b, &c, &d);
      double l1 = b - a + 1, l2 = d - c + 1;
      double s1 = query(1, 1, n, a, b);
      double s2 = query(1, 1, n, c, d);
      mul_qj(1, 1, n, a, b, (l1 - 1) / l1);
      mul_qj(1, 1, n, c, d, (l2 - 1) / l2);
      add_qj(1, 1, n, a, b, 1.0 / l1 * (s2 / l2));
      add_qj(1, 1, n, c, d, 1.0 / l2 * (s1 / l1));
    } else {
      int a, b;
      scanf("%d%d", &a, &b);
      printf("%.5f\n", query(1, 1, n, a, b));
    }
  }
  return 0;
}