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 N = 4e5 + 10;
double add[N], mul[N], s[N];
int n, q;
void pushup(int i) { s[i] = s[(i << 1)] + s[(i << 1 | 1)]; }
void pushdown(int i, int L, int R) {
  s[(i << 1)] = mul[i] * s[(i << 1)] + add[i] * (((L + R) >> 1) - L + 1);
  mul[(i << 1)] *= mul[i];
  add[(i << 1)] = mul[i] * add[(i << 1)] + add[i];
  s[(i << 1 | 1)] = mul[i] * s[(i << 1 | 1)] + add[i] * (R - ((L + R) >> 1));
  mul[(i << 1 | 1)] *= mul[i];
  add[(i << 1 | 1)] = mul[i] * add[(i << 1 | 1)] + add[i];
  add[i] = 0;
  mul[i] = 1.0;
}
void build(int i, int L, int R) {
  add[i] = 0, mul[i] = 1;
  s[i] = 0;
  if (L == R) {
    scanf("%lf", s + i);
    return;
  }
  build(i << 1, L, ((L + R) >> 1));
  build(i << 1 | 1, ((L + R) >> 1) + 1, R);
  pushup(i);
}
void add_update(int i, int L, int R, int l, int r, double val) {
  if (l <= L && R <= r) {
    s[i] += (R - L + 1) * val;
    add[i] += val;
    return;
  }
  pushdown(i, L, R);
  if (l <= ((L + R) >> 1)) add_update(i << 1, L, ((L + R) >> 1), l, r, val);
  if (r > ((L + R) >> 1))
    add_update(i << 1 | 1, ((L + R) >> 1) + 1, R, l, r, val);
  pushup(i);
}
void mul_update(int i, int L, int R, int l, int r, double val) {
  if (l <= L && R <= r) {
    s[i] *= val;
    mul[i] *= val;
    add[i] *= val;
    return;
  }
  pushdown(i, L, R);
  if (l <= ((L + R) >> 1)) mul_update(i << 1, L, ((L + R) >> 1), l, r, val);
  if (r > ((L + R) >> 1))
    mul_update(i << 1 | 1, ((L + R) >> 1) + 1, R, l, r, val);
  pushup(i);
}
double query(int i, int L, int R, int l, int r) {
  if (l <= L && R <= r) return s[i];
  double ret = 0;
  pushdown(i, L, R);
  if (l <= ((L + R) >> 1)) ret += query(i << 1, L, ((L + R) >> 1), l, r);
  if (r > ((L + R) >> 1)) ret += query(i << 1 | 1, ((L + R) >> 1) + 1, R, l, r);
  pushup(i);
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  build(1, 1, n);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l1, r1, l2, r2;
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double s1 = query(1, 1, n, l1, r1);
      double s2 = query(1, 1, n, l2, r2);
      double len1 = r1 - l1 + 1;
      double len2 = r2 - l2 + 1;
      mul_update(1, 1, n, l1, r1, 1.0 * (len1 - 1) / len1);
      mul_update(1, 1, n, l2, r2, 1.0 * (len2 - 1) / len2);
      add_update(1, 1, n, l1, r1, 1.0 / len1 / len2 * s2);
      add_update(1, 1, n, l2, r2, 1.0 / len1 / len2 * s1);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%.12f\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}