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 a[MAXN];
double tr[MAXN << 2], multag[MAXN << 2], addtag[MAXN << 2];
void pushUp(int rt) { tr[rt] = tr[rt << 1] + tr[rt << 1 | 1]; }
void buildTree(int rt, int l, int r) {
  addtag[rt] = 0, multag[rt] = 1.0;
  if (l == r) {
    tr[rt] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  buildTree(rt << 1, l, mid), buildTree(rt << 1 | 1, mid + 1, r);
  pushUp(rt);
}
void pushDownMul(int rt, int l, int r) {
  if (l < r) {
    tr[rt << 1] *= multag[rt], addtag[rt << 1] *= multag[rt],
        multag[rt << 1] *= multag[rt];
    tr[rt << 1 | 1] *= multag[rt], addtag[rt << 1 | 1] *= multag[rt],
        multag[rt << 1 | 1] *= multag[rt];
  }
  multag[rt] = 1.0;
}
void pushDownAdd(int rt, int l, int r) {
  if (l < r) {
    int mid = (l + r) / 2;
    tr[rt << 1] += (mid - l + 1) * addtag[rt], addtag[rt << 1] += addtag[rt];
    tr[rt << 1 | 1] += (r - mid) * addtag[rt],
        addtag[rt << 1 | 1] += addtag[rt];
  }
  addtag[rt] = 0;
}
void modifyAdd(int rt, int l, int r, int s, int t, double x) {
  if (multag[rt] != 1.0) pushDownMul(rt, l, r);
  if (addtag[rt]) pushDownAdd(rt, l, r);
  if (s <= l && t >= r) {
    tr[rt] += x * (r - l + 1), addtag[rt] += x;
    return;
  }
  int mid = (l + r) / 2;
  if (s <= mid) modifyAdd(rt << 1, l, mid, s, t, x);
  if (t > mid) modifyAdd(rt << 1 | 1, mid + 1, r, s, t, x);
  pushUp(rt);
}
void modifyMul(int rt, int l, int r, int s, int t, double x) {
  if (multag[rt] != 1.0) pushDownMul(rt, l, r);
  if (addtag[rt]) pushDownAdd(rt, l, r);
  if (s <= l && t >= r) {
    tr[rt] *= x, addtag[rt] *= x, multag[rt] *= x;
    return;
  }
  int mid = (l + r) / 2;
  if (s <= mid) modifyMul(rt << 1, l, mid, s, t, x);
  if (t > mid) modifyMul(rt << 1 | 1, mid + 1, r, s, t, x);
  pushUp(rt);
}
double query(int rt, int l, int r, int s, int t) {
  if (multag[rt] != 1.0) pushDownMul(rt, l, r);
  if (addtag[rt]) pushDownAdd(rt, l, r);
  if (s <= l && t >= r) return tr[rt];
  int mid = (l + r) / 2;
  double ans = 0.0;
  if (s <= mid) ans = query(rt << 1, l, mid, s, t);
  if (t > mid) ans += query(rt << 1 | 1, mid + 1, r, s, t);
  pushUp(rt);
  return ans;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  buildTree(1, 1, n);
  int k, l1, r1, l2, r2;
  for (int i = 1; i <= q; i++) {
    scanf("%d", &k);
    if (k == 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);
      int a = r1 - l1 + 1, b = r2 - l2 + 1;
      modifyMul(1, 1, n, l1, r1, 1.0 * (a - 1) / a),
          modifyAdd(1, 1, n, l1, r1, 1.0 * sum2 / a / b);
      modifyMul(1, 1, n, l2, r2, 1.0 * (b - 1) / b),
          modifyAdd(1, 1, n, l2, r2, 1.0 * sum1 / a / b);
    } else {
      scanf("%d%d", &l1, &r1);
      printf("%lf\n", query(1, 1, n, l1, r1));
    }
  }
}