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 long double PI = acos(-1);
const long double eps = 1e-9;
const int seed = 131;
const int M = 1e5 + 7;
double a[M];
double sm[M << 2];
double mul[M << 2];
double add[M << 2];
void bd(int o, int l, int r) {
  mul[o] = 1, add[o] = 0;
  if (l == r) {
    mul[o] = 1;
    add[o] = 0;
    sm[o] = a[l];
    return;
  }
  int m = (l + r) / 2;
  bd((o << 1), l, m);
  bd((o << 1 | 1), m + 1, r);
  sm[o] = sm[(o << 1)] + sm[(o << 1 | 1)];
}
void pd(int o, int l, int r) {
  int m = (l + r) / 2;
  {
    mul[(o << 1)] *= mul[o], mul[(o << 1 | 1)] *= mul[o];
    add[(o << 1)] *= mul[o], add[(o << 1 | 1)] *= mul[o];
    add[(o << 1)] += add[o], add[(o << 1 | 1)] += add[o];
    sm[(o << 1)] = sm[(o << 1)] * mul[o] + add[o] * (m - l + 1);
    sm[(o << 1 | 1)] = sm[(o << 1 | 1)] * mul[o] + add[o] * (r - m);
    mul[o] = 1;
    add[o] = 0;
  }
}
void up(int o, int l, int r, int x, int y, double k, double b) {
  if (x <= l && r <= y) {
    mul[o] *= k, add[o] *= k, add[o] += b;
    sm[o] = sm[o] * k + b * (r - l + 1);
    return;
  }
  pd(o, l, r);
  int m = (l + r) / 2;
  if (x <= m) up((o << 1), l, m, x, y, k, b);
  if (y > m) up((o << 1 | 1), m + 1, r, x, y, k, b);
  sm[o] = sm[(o << 1)] + sm[(o << 1 | 1)];
}
double qu(int o, int l, int r, int x, int y) {
  if (x <= l && r <= y) {
    return sm[o];
  }
  int m = (l + r) / 2;
  pd(o, l, r);
  double ans = 0;
  if (x <= m) ans += qu((o << 1), l, m, x, y);
  if (y > m) ans += qu((o << 1 | 1), m + 1, r, x, y);
  return ans;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%lf", &a[i]);
  bd(1, 1, n);
  while (q--) {
    int opt, l1, r1, l2, r2;
    double sm1 = 0, sm2 = 0, k, b;
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      sm1 = qu(1, 1, n, l1, r1);
      sm2 = qu(1, 1, n, l2, r2);
      k = 1.0 * (r1 - l1) / (r1 - l1 + 1);
      b = sm2 / (r2 - l2 + 1);
      b /= (r1 - l1 + 1);
      up(1, 1, n, l1, r1, k, b);
      k = 1.0 * (r2 - l2) / (r2 - l2 + 1);
      b = sm1 / (r1 - l1 + 1);
      b /= (r2 - l2 + 1);
      up(1, 1, n, l2, r2, k, b);
    } else {
      scanf("%d%d", &l1, &r1);
      double tp = qu(1, 1, n, l1, r1);
      printf("%.7f\n", tp);
    }
  }
  return 0;
}