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;
double arr[200200];
struct matr22 {
  double a, b, c, d;
  bool operator==(const matr22& x) {
    return x.a == a && x.b == b && x.c == c && x.d == d;
  }
};
matr22 multMatr(matr22 A, matr22 B) {
  matr22 toRet = {A.a * B.a + A.b * B.c, A.a * B.b + A.b * B.d,
                  A.c * B.a + A.d * B.c, A.c * B.b + A.d * B.d};
  return toRet;
}
matr22 idMatr() {
  matr22 toRet = {1, 0, 0, 1};
  return toRet;
}
struct nodeS {
  int l, r;
  nodeS *nodeL, *nodeR;
  double S;
  matr22 lazyMatr;
};
nodeS* buildSegmentTree(int l, int r) {
  nodeS* node = new nodeS();
  node->nodeL = node->nodeR = NULL;
  node->l = l;
  node->r = r;
  node->lazyMatr = idMatr();
  if (l == r) {
    node->S = arr[l];
    return node;
  } else {
    int m = (l + r) / 2;
    node->nodeL = buildSegmentTree(l, m);
    node->nodeR = buildSegmentTree(m + 1, r);
    node->S = (node->nodeL->S + node->nodeR->S);
    return node;
  }
}
void pushLazy(nodeS* now) {
  if (!(now->lazyMatr == idMatr())) {
    now->S = now->S * now->lazyMatr.a + (now->r - now->l + 1) * now->lazyMatr.b;
    if (now->nodeL != NULL) {
      now->nodeL->lazyMatr = multMatr(now->lazyMatr, now->nodeL->lazyMatr);
      now->nodeR->lazyMatr = multMatr(now->lazyMatr, now->nodeR->lazyMatr);
    }
    now->lazyMatr = idMatr();
  }
}
double query(nodeS* now, int l, int r) {
  if (now->r < l || now->l > r) return 0;
  pushLazy(now);
  if (l <= now->l && now->r <= r) return now->S;
  return query(now->nodeL, l, r) + query(now->nodeR, l, r);
}
void update(nodeS* now, int l, int r, double L, double y) {
  if (now->r < l || now->l > r) return;
  pushLazy(now);
  if (l <= now->l && now->r <= r) {
    matr22 mult = {(L - 1) / L, y / L, 0, 1};
    now->lazyMatr = multMatr(mult, now->lazyMatr);
  } else {
    update(now->nodeL, l, r, L, y);
    update(now->nodeR, l, r, L, y);
    now->S = query(now->nodeL, now->nodeL->l, now->nodeL->r) +
             query(now->nodeR, now->nodeR->l, now->nodeR->r);
  }
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%lf", &arr[i]);
  }
  nodeS* root = buildSegmentTree(0, n - 1);
  for (int i = 0; i < q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int l, r, l2, r2;
      scanf("%d %d %d %d", &l, &r, &l2, &r2);
      l--;
      r--;
      l2--;
      r2--;
      double L1 = r - l + 1;
      double L2 = r2 - l2 + 1;
      double x = query(root, l, r) / L1;
      double y = query(root, l2, r2) / L2;
      update(root, l, r, r - l + 1, y);
      update(root, l2, r2, r2 - l2 + 1, x);
    } else if (t == 3) {
      for (int j = 0; j < n; j++) printf("%f ", query(root, j, j));
      printf("\n");
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      l--;
      r--;
      printf("%f\n", query(root, l, r));
    }
  }
}