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 = 100010;
int n, q, L[maxn << 2], R[maxn << 2];
double sum[maxn << 2], add[maxn << 2], mul[maxn << 2];
void pushup(int o) { sum[o] = sum[o * 2] + sum[o * 2 + 1]; }
void pushdown(int o) {
  sum[o * 2] = sum[o * 2] * mul[o] + add[o] * (R[o * 2] - L[o * 2] + 1);
  sum[o * 2 + 1] =
      sum[o * 2 + 1] * mul[o] + add[o] * (R[o * 2 + 1] - L[o * 2 + 1] + 1);
  add[o * 2] = add[o * 2] * mul[o] + add[o];
  add[o * 2 + 1] = add[o * 2 + 1] * mul[o] + add[o];
  mul[o * 2] = mul[o * 2] * mul[o];
  mul[o * 2 + 1] = mul[o * 2 + 1] * mul[o];
  mul[o] = 1;
  add[o] = 0;
}
void build(int o, int l, int r) {
  L[o] = l;
  R[o] = r;
  add[o] = 0;
  mul[o] = 1;
  if (l == r) {
    cin >> sum[o];
    return;
  }
  int mid = (l + r) >> 1;
  build(o * 2, l, mid);
  build(o * 2 + 1, mid + 1, r);
  pushup(o);
}
double query(int o, int l, int r, int x, int y) {
  if (x <= l && r <= y) {
    return sum[o];
  }
  pushdown(o);
  int mid = (l + r) >> 1;
  double res = 0.0;
  if (x <= mid) res += query(o * 2, l, mid, x, y);
  if (y > mid) res += query(o * 2 + 1, mid + 1, r, x, y);
  return res;
}
void update1(int o, int l, int r, int x, int y, double v) {
  if (x <= l && r <= y) {
    sum[o] += v * (r - l + 1);
    add[o] += v;
    return;
  }
  pushdown(o);
  int mid = (l + r) >> 1;
  if (x <= mid) update1(o * 2, l, mid, x, y, v);
  if (y > mid) update1(o * 2 + 1, mid + 1, r, x, y, v);
  pushup(o);
}
void update2(int o, int l, int r, int x, int y, double v) {
  if (x <= l && r <= y) {
    add[o] *= v;
    sum[o] *= v;
    mul[o] *= v;
    return;
  }
  pushdown(o);
  int mid = (l + r) >> 1;
  if (x <= mid) update2(o * 2, l, mid, x, y, v);
  if (y > mid) update2(o * 2 + 1, mid + 1, r, x, y, v);
  pushup(o);
}
int main() {
  scanf("%d%d", &n, &q);
  build(1, 1, n);
  while (q--) {
    int l1, r1, l2, r2;
    int id;
    scanf("%d", &id);
    if (id == 1) {
      scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
      double sum1 = query(1, 1, n, l1, r1), sum2 = query(1, 1, n, l2, r2);
      double len1 = r1 - l1 + 1, len2 = r2 - l2 + 1;
      update2(1, 1, n, l1, r1, (len1 - 1) / len1);
      update2(1, 1, n, l2, r2, (len2 - 1) / len2);
      update1(1, 1, n, l1, r1, 1.0 / len1 * (sum2 / len2));
      update1(1, 1, n, l2, r2, 1.0 / len2 * (sum1 / len1));
    } else {
      scanf("%d%d", &l1, &r1);
      printf("%.7lf\n", query(1, 1, n, l1, r1));
    }
  }
  return 0;
}