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 = 1e5 + 5;
const double eps = 1e-10;
double node[maxn << 2], lazy[maxn << 2], mul[maxn << 2];
int dcmp(double x) {
  if (fabs(x) < eps) return 0;
  return x > 0 ? 1 : -1;
}
void modify(int rt, int l, int r, double c) {
  lazy[rt] += c;
  node[rt] += c * (r - l + 1);
}
void modify1(int rt, double c) {
  mul[rt] *= c;
  lazy[rt] *= c;
  node[rt] *= c;
}
void pushup(int rt) { node[rt] = node[rt << 1] + node[rt << 1 | 1]; }
void pushdown(int rt, int l, int r) {
  if (dcmp(mul[rt] - 1)) {
    modify1(rt << 1, mul[rt]);
    modify1(rt << 1 | 1, mul[rt]);
    mul[rt] = 1;
  }
  if (dcmp(lazy[rt])) {
    int m = (l + r) >> 1;
    modify(rt << 1, l, m, lazy[rt]);
    modify(rt << 1 | 1, m + 1, r, lazy[rt]);
    lazy[rt] = 0;
  }
}
void build(int rt, int l, int r) {
  lazy[rt] = 0;
  mul[rt] = 1;
  if (l == r) {
    int a;
    scanf("%d", &a);
    node[rt] = a;
    return;
  }
  int m = (l + r) >> 1;
  build(rt << 1, l, m);
  build(rt << 1 | 1, m + 1, r);
  pushup(rt);
}
void update(int rt, int l, int r, int x, int y, double c) {
  if (x <= l && r <= y) {
    modify(rt, l, r, c);
    return;
  }
  pushdown(rt, l, r);
  int m = (l + r) >> 1;
  if (x <= m) update(rt << 1, l, m, x, y, c);
  if (m < y) update(rt << 1 | 1, m + 1, r, x, y, c);
  pushup(rt);
}
void update1(int rt, int l, int r, int x, int y, double c) {
  if (x <= l && r <= y) {
    modify1(rt, c);
    return;
  }
  pushdown(rt, l, r);
  int m = (l + r) >> 1;
  if (x <= m) update1(rt << 1, l, m, x, y, c);
  if (m < y) update1(rt << 1 | 1, m + 1, r, x, y, c);
  pushup(rt);
}
double query(int rt, int l, int r, int x, int y) {
  if (x <= l && r <= y) return node[rt];
  pushdown(rt, l, r);
  int m = (l + r) >> 1;
  double ret = 0;
  if (x <= m) ret += query(rt << 1, l, m, x, y);
  if (m < y) ret += query(rt << 1 | 1, m + 1, r, x, y);
  return ret;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  build(1, 1, n);
  while (q--) {
    int op, l1, r1, l2, r2;
    scanf("%d%d%d", &op, &l1, &r1);
    if (op == 1) {
      scanf("%d%d", &l2, &r2);
      double c1 = query(1, 1, n, l1, r1) / (r1 - l1 + 1),
             c2 = query(1, 1, n, l2, r2) / (r2 - l2 + 1);
      update1(1, 1, n, l1, r1, (double)(r1 - l1) / (r1 - l1 + 1));
      update(1, 1, n, l1, r1, c2 / (r1 - l1 + 1));
      update1(1, 1, n, l2, r2, (double)(r2 - l2) / (r2 - l2 + 1));
      update(1, 1, n, l2, r2, c1 / (r2 - l2 + 1));
    } else
      printf("%.10f\n", query(1, 1, n, l1, r1));
  }
  return 0;
}