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;
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int &x, int y, int mod = 1000000007) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et() {
  puts("-1");
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) {
  long long z = y;
  while (x % y != 0) {
    z = x % y;
    x = y;
    y = z;
  }
  return z;
}
int a[100005];
const int MN = 100005 << 2;
long double T[MN], f[MN], b[MN];
bool lz[MN];
void Up(int rt, int l, int r) { T[rt] = T[(rt << 1)] + T[(rt << 1 | 1)]; }
void build(int rt, int l, int r) {
  f[rt] = 1;
  if (l + 1 == r) {
    T[rt] = a[r];
    return;
  }
  int mid = (l + r) / 2;
  build((rt << 1), l, mid);
  build((rt << 1 | 1), mid, r);
  Up(rt, l, r);
}
void Down(int rt, int l, int r) {
  if (lz[rt]) {
    int mid = (l + r) / 2;
    f[(rt << 1)] = f[(rt << 1)] * f[rt];
    b[(rt << 1)] = f[rt] * b[(rt << 1)] + b[rt];
    T[(rt << 1)] = T[(rt << 1)] * f[rt] + b[rt] * (mid - l);
    lz[(rt << 1)] = 1;
    f[(rt << 1 | 1)] = f[(rt << 1 | 1)] * f[rt];
    b[(rt << 1 | 1)] = f[rt] * b[(rt << 1 | 1)] + b[rt];
    T[(rt << 1 | 1)] = T[(rt << 1 | 1)] * f[rt] + b[rt] * (r - mid);
    lz[(rt << 1 | 1)] = 1;
    lz[rt] = 0;
    f[rt] = 1;
    b[rt] = 0;
  }
}
void upt(int rt, int l, int r, int L, int R, long double K, long double B) {
  if (R <= l || r <= L) return;
  if (L <= l && r <= R) {
    f[rt] = f[rt] * K;
    b[rt] = K * b[rt] + B;
    T[rt] = T[rt] * K + B * (r - l);
    lz[rt] = 1;
    return;
  }
  Down(rt, l, r);
  int mid = (l + r) / 2;
  upt((rt << 1), l, mid, L, R, K, B);
  upt((rt << 1 | 1), mid, r, L, R, K, B);
  Up(rt, l, r);
}
long double qy(int rt, int l, int r, int L, int R) {
  if (R <= l || r <= L) return 0;
  if (L <= l && r <= R) {
    return T[rt];
  }
  Down(rt, l, r);
  int mid = (l + r) / 2;
  return qy((rt << 1), l, mid, L, R) + qy((rt << 1 | 1), mid, r, L, R);
}
void fmain(int ID) {
  scanf("%d%d", &n, &m);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%d", a + i);
  build(1, 0, n);
  for (int(i) = 1; (i) <= (int)(m); (i)++) {
    int t, l, r, L, R;
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) {
      scanf("%d%d", &L, &R);
      long double v1 = qy(1, 0, n, l - 1, r);
      long double v2 = qy(1, 0, n, L - 1, R);
      int c1 = r - l + 1, c2 = R - L + 1;
      upt(1, 0, n, l - 1, r, (long double)(c1 - 1) / c1, v2 / c1 / c2);
      upt(1, 0, n, L - 1, R, (long double)(c2 - 1) / c2, v1 / c1 / c2);
    } else {
      printf("%.10Lf\n", qy(1, 0, n, l - 1, r));
    }
  }
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}