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;
namespace IO {
template <class T>
inline bool read(T &x) {
  int c = getchar();
  int sgn = 1;
  while (~c && c < '0' || c > '9') {
    if (c == '-') sgn = -1;
    c = getchar();
  }
  for (x = 0; ~c && '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= sgn;
  return ~c;
}
}  // namespace IO
namespace Utility {
template <typename T>
T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <typename T>
T lcm(T a, T b) {
  return (a / gcd(a, b) * b);
}
template <typename T>
inline T addmod(T a, T b, T mod) {
  T c = a + b;
  return c > mod ? c - mod : c;
}
template <typename T>
inline T submod(T a, T b, T mod) {
  T c = a - b;
  return c < mod ? c + mod : c;
}
template <typename T>
inline T mulmod(T a, T b, T mod) {
  return a * 1LL * b % mod;
}
template <typename T>
T exEuclid(T a, T b, T &x, T &y) {
  x = 0, y = 1;
  if (a == 0) return b;
  T x1, y1;
  T d = exEuclid(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
template <typename T>
T modInv(T a, T mod) {
  T x, y;
  exEuclid(a, mod, x, y);
  return x < 0 ? x + mod : x;
}
template <typename T>
T bigmod(T b, T p, T mod) {
  T ret = b > 0;
  while (p) {
    if (p & 1) ret = mulmod(ret, b, mod);
    p >>= 1, b = mulmod(b, b, mod);
  }
  return ret;
}
}  // namespace Utility
using namespace IO;
using namespace Utility;
const int INF = 2000000009;
const int MX = 400005;
const double EPS = 1e-9;
const double PI = acos(-1.0);
const int MOD = 1000000007;
double a[MX], sum[MX], sum2[MX], prod[MX];
void build(int node, int l, int r) {
  sum2[node] = 0;
  prod[node] = 1;
  if (l == r) {
    sum[node] = a[l];
    return;
  }
  int mid = l + r >> 1;
  build(node << 1, l, mid);
  build(node << 1 | 1, mid + 1, r);
  sum[node] = sum[node << 1] + sum[node << 1 | 1];
}
void lazyUpdate(int node, int l, int r) {
  sum[node] = sum[node] * prod[node] + sum2[node] * (r - l + 1);
  if (l != r) {
    prod[node << 1] *= prod[node];
    sum2[node << 1] = sum2[node << 1] * prod[node] + sum2[node];
    prod[node << 1 | 1] *= prod[node];
    sum2[node << 1 | 1] = sum2[node << 1 | 1] * prod[node] + sum2[node];
  }
  prod[node] = 1;
  sum2[node] = 0;
}
void update(int node, int l, int r, int i, int j, double mul, double add) {
  lazyUpdate(node, l, r);
  if (l > j || r < i) return;
  if (l >= i && r <= j) {
    prod[node] *= mul;
    sum2[node] = sum2[node] * mul + add;
    lazyUpdate(node, l, r);
    return;
  }
  int mid = l + r >> 1;
  update(node << 1, l, mid, i, j, mul, add);
  update(node << 1 | 1, mid + 1, r, i, j, mul, add);
  sum[node] = sum[node << 1] + sum[node << 1 | 1];
}
double query(int node, int l, int r, int i, int j) {
  lazyUpdate(node, l, r);
  if (l > j || r < i) return 0;
  if (l >= i && r <= j) return sum[node];
  int mid = l + r >> 1;
  return query(node << 1, l, mid, i, j) +
         query(node << 1 | 1, mid + 1, r, i, j);
}
int main() {
  int n = ({
    int a;
    read(a);
    a;
  }),
      q = ({
        int a;
        read(a);
        a;
      });
  for (int i = 1; i <= n; i++)
    a[i] = ({
      int a;
      read(a);
      a;
    });
  build(1, 1, n);
  while (q--) {
    int t = ({
      int a;
      read(a);
      a;
    });
    if (t == 1) {
      int l1 = ({
        int a;
        read(a);
        a;
      }),
          r1 = ({
            int a;
            read(a);
            a;
          }),
          l2 = ({
            int a;
            read(a);
            a;
          }),
          r2 = ({
            int a;
            read(a);
            a;
          });
      double mul1 = (1.0 * r1 - l1) / (r1 - l1 + 1);
      double add1 = query(1, 1, n, l2, r2) / (r2 - l2 + 1) / (r1 - l1 + 1);
      double mul2 = (1.0 * r2 - l2) / (r2 - l2 + 1);
      double add2 = query(1, 1, n, l1, r1) / (r1 - l1 + 1) / (r2 - l2 + 1);
      update(1, 1, n, l1, r1, mul1, add1);
      update(1, 1, n, l2, r2, mul2, add2);
    } else {
      int l = ({
        int a;
        read(a);
        a;
      }),
          r = ({
            int a;
            read(a);
            a;
          });
      printf("%.7f\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}