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;
const int mod = 1e9 + 7;
const int MAX = 1000000010;
const double eps = 1e-6;
const double PI = acos(-1);
template <typename T>
inline void read(T &x) {
  x = 0;
  T f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do x = x * 10 + ch - '0', ch = getchar();
  while (ch <= '9' && ch >= '0');
  x *= f;
}
template <typename A, typename B>
inline void read(A &x, B &y) {
  read(x);
  read(y);
}
template <typename A, typename B, typename C>
inline void read(A &x, B &y, C &z) {
  read(x);
  read(y);
  read(z);
}
template <typename A, typename B, typename C, typename D>
inline void read(A &x, B &y, C &z, D &w) {
  read(x);
  read(y);
  read(z);
  read(w);
}
template <typename A, typename B>
inline A fexp(A x, B p) {
  A ans = 1;
  for (; p; p >>= 1, x = 1LL * x * x % mod)
    if (p & 1) ans = 1LL * ans * x % mod;
  return ans;
}
template <typename A, typename B>
inline A fexp(A x, B p, A mo) {
  A ans = 1;
  for (; p; p >>= 1, x = 1LL * x * x % mo)
    if (p & 1) ans = 1LL * ans * x % mo;
  return ans;
}
int n, m;
double A[maxn];
struct Node {
  int l, r;
  double sm, bjc, bjj;
} T[maxn << 2];
void pushup(int i) { T[i].sm = T[i << 1].sm + T[i << 1 | 1].sm; }
void mul(int i, double x) {
  T[i].sm *= x;
  T[i].bjc *= x;
  T[i].bjj *= x;
}
void add(int i, double x) {
  T[i].sm += x * (T[i].r - T[i].l + 1);
  T[i].bjj += x;
}
void pushdown(int i) {
  mul(i << 1, T[i].bjc);
  mul(i << 1 | 1, T[i].bjc);
  T[i].bjc = 1;
  add(i << 1, T[i].bjj);
  add(i << 1 | 1, T[i].bjj);
  T[i].bjj = 0;
}
void build(int i, int l, int r) {
  T[i].l = l;
  T[i].r = r;
  T[i].bjc = 1;
  if (l == r) {
    T[i].sm = A[l];
    return;
  }
  int M = l + r >> 1;
  build(i << 1, l, M);
  build(i << 1 | 1, M + 1, r);
  pushup(i);
}
double query(int i, int l, int r) {
  if (l <= T[i].l && T[i].r <= r) return T[i].sm;
  int M = T[i].l + T[i].r >> 1;
  double ans = 0;
  pushdown(i);
  if (l <= M) ans += query(i << 1, l, r);
  if (r > M) ans += query(i << 1 | 1, l, r);
  return ans;
}
void add(int i, int l, int r, double x) {
  if (l <= T[i].l && T[i].r <= r) {
    add(i, x);
    return;
  }
  int M = T[i].l + T[i].r >> 1;
  pushdown(i);
  if (l <= M) add(i << 1, l, r, x);
  if (r > M) add(i << 1 | 1, l, r, x);
  pushup(i);
}
void mul(int i, int l, int r, double x) {
  if (l <= T[i].l && T[i].r <= r) {
    mul(i, x);
    return;
  }
  int M = T[i].l + T[i].r >> 1;
  pushdown(i);
  if (l <= M) mul(i << 1, l, r, x);
  if (r > M) mul(i << 1 | 1, l, r, x);
  pushup(i);
}
int main() {
  read(n, m);
  for (int i = (1); i <= (n); i++) scanf("%lf", &A[i]);
  build(1, 1, n);
  while (m--) {
    int tp;
    read(tp);
    if (tp == 1) {
      int l1, r1, l2, r2;
      read(l1, r1, l2, r2);
      double sm1 = 0, sm2 = 0;
      sm1 = query(1, l1, r1);
      sm2 = query(1, l2, r2);
      mul(1, l1, r1, (double)(r1 - l1) / (r1 - l1 + 1));
      mul(1, l2, r2, (double)(r2 - l2) / (r2 - l2 + 1));
      add(1, l1, r1, sm2 / (r2 - l2 + 1) / (r1 - l1 + 1));
      add(1, l2, r2, sm1 / (r2 - l2 + 1) / (r1 - l1 + 1));
    } else {
      int l, r;
      read(l, r);
      printf("%.10lf\n", query(1, l, r));
    }
  }
  return 0;
}