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>
inline int read() {
  char c = getchar();
  int x = 0;
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  return x;
}
const int N = 2e5 + 5;
int n, q, a[N];
int rt = 0, node_cnt = 0, ch[N][2], ql, qr;
double sum[N], tgp[N], tgm[N], qc;
void pushdown(int l, int r, int u) {
  int mid = l + r >> 1;
  sum[ch[u][0]] *= tgm[u];
  tgp[ch[u][0]] *= tgm[u];
  tgm[ch[u][0]] *= tgm[u];
  sum[ch[u][0]] += tgp[u] * (mid - l + 1);
  tgp[ch[u][0]] += tgp[u];
  sum[ch[u][1]] *= tgm[u];
  tgp[ch[u][1]] *= tgm[u];
  tgm[ch[u][1]] *= tgm[u];
  sum[ch[u][1]] += tgp[u] * (r - mid);
  tgp[ch[u][1]] += tgp[u];
  tgp[u] = 0.0;
  tgm[u] = 1.0;
}
void build(int l, int r, int& u) {
  u = ++node_cnt;
  tgm[u] = 1.0;
  if (l == r) {
    sum[u] = (double)a[l];
  } else {
    int mid = l + r >> 1;
    build(l, mid, ch[u][0]);
    build(mid + 1, r, ch[u][1]);
    sum[u] = sum[ch[u][0]] + sum[ch[u][1]];
  }
}
void update(int l, int r, int u) {
  if (ql <= l && r <= qr) {
    sum[u] += qc * (r - l + 1);
    tgp[u] += qc;
  } else {
    int mid = l + r >> 1;
    pushdown(l, r, u);
    if (ql <= mid) update(l, mid, ch[u][0]);
    if (qr > mid) update(mid + 1, r, ch[u][1]);
    sum[u] = sum[ch[u][0]] + sum[ch[u][1]];
  }
}
void modify(int l, int r, int u) {
  if (ql <= l && r <= qr) {
    sum[u] *= qc;
    tgp[u] *= qc;
    tgm[u] *= qc;
  } else {
    int mid = l + r >> 1;
    pushdown(l, r, u);
    if (ql <= mid) modify(l, mid, ch[u][0]);
    if (qr > mid) modify(mid + 1, r, ch[u][1]);
    sum[u] = sum[ch[u][0]] + sum[ch[u][1]];
  }
}
double query(int l, int r, int u) {
  if (ql <= l && r <= qr) {
    return sum[u];
  } else {
    int mid = l + r >> 1;
    pushdown(l, r, u);
    if (qr <= mid) return query(l, mid, ch[u][0]);
    if (ql > mid) return query(mid + 1, r, ch[u][1]);
    return query(l, mid, ch[u][0]) + query(mid + 1, r, ch[u][1]);
  }
}
int main() {
  n = read();
  q = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  build(1, n, rt);
  for (int i = 1, opt, len1, len2, l1, r1, l2, r2; i <= q; ++i) {
    if ((opt = read()) == 1) {
      l1 = read();
      r1 = read();
      l2 = read();
      r2 = read();
      ql = l1;
      qr = r1;
      double s1 = query(1, n, rt);
      len1 = r1 - l1 + 1;
      ql = l2;
      qr = r2;
      double s2 = query(1, n, rt);
      len2 = r2 - l2 + 1;
      ql = l1;
      qr = r1;
      qc = (double)(len1 - 1) / len1;
      modify(1, n, rt);
      ql = l1;
      qr = r1;
      qc = (double)s2 / len1 / len2;
      update(1, n, rt);
      ql = l2;
      qr = r2;
      qc = (double)(len2 - 1) / len2;
      modify(1, n, rt);
      ql = l2;
      qr = r2;
      qc = (double)s1 / len1 / len2;
      update(1, n, rt);
    } else {
      ql = read();
      qr = read();
      printf("%.7lf\n", query(1, n, rt));
    }
  }
  return 0;
}