You are given an array a of length 2^n. You should process q queries on it. Each query has one of the following 4 types: 

  1. Replace(x, k) — change a_x to k; 
  2. Reverse(k) — reverse each subarray [(i-1) ⋅ 2^k+1, i ⋅ 2^k] for all i (i ≥ 1); 
  3. Swap(k) — swap subarrays [(2i-2) ⋅ 2^k+1, (2i-1) ⋅ 2^k] and [(2i-1) ⋅ 2^k+1, 2i ⋅ 2^k] for all i (i ≥ 1); 
  4. Sum(l, r) — print the sum of the elements of subarray [l, r]. 



Write a program that can quickly process given queries.

Input

The first line contains two integers n, q (0 ≤ n ≤ 18; 1 ≤ q ≤ 10^5) — the length of array a and the number of queries.

The second line contains 2^n integers a_1, a_2, …, a_{2^n} (0 ≤ a_i ≤ 10^9).

Next q lines contains queries — one per line. Each query has one of 4 types: 

  * "1 x k" (1 ≤ x ≤ 2^n; 0 ≤ k ≤ 10^9) — Replace(x, k); 
  * "2 k" (0 ≤ k ≤ n) — Reverse(k); 
  * "3 k" (0 ≤ k < n) — Swap(k); 
  * "4 l r" (1 ≤ l ≤ r ≤ 2^n) — Sum(l, r). 



It is guaranteed that there is at least one Sum query.

Output

Print the answer for each Sum query.

Examples

Input


2 3
7 4 9 9
1 2 8
3 1
4 2 4


Output


24


Input


3 8
7 0 8 8 7 1 5 2
4 3 7
2 1
3 2
4 1 6
2 3
1 5 16
4 8 8
3 0


Output


29
22
1

Note

In the first sample, initially, the array a is equal to \{7,4,9,9\}.

After processing the first query. the array a becomes \{7,8,9,9\}.

After processing the second query, the array a_i becomes \{9,9,7,8\}

Therefore, the answer to the third query is 9+7+8=24.

In the second sample, initially, the array a is equal to \{7,0,8,8,7,1,5,2\}. What happens next is: 

  1. Sum(3, 7) → 8 + 8 + 7 + 1 + 5 = 29; 
  2. Reverse(1) → \{0,7,8,8,1,7,2,5\}; 
  3. Swap(2) → \{1,7,2,5,0,7,8,8\}; 
  4. Sum(1, 6) → 1 + 7 + 2 + 5 + 0 + 7 = 22; 
  5. Reverse(3) → \{8,8,7,0,5,2,7,1\}; 
  6. Replace(5, 16) → \{8,8,7,0,16,2,7,1\}; 
  7. Sum(8, 8) → 1; 
  8. Swap(0) → \{8,8,0,7,2,16,1,7\}. 
#include <bits/stdc++.h>
using namespace std;
int n, q;
int kek[20];
long long a[696969];
void upd(int x, int v, int s = 0, int e = (1 << n) - 1, int d = n - 1,
         int p = 1) {
  if (s == e) {
    a[p] = v;
    return;
  }
  int m = (s + e) / 2;
  if (x <= m) {
    upd(x, v, s, m, d - 1, 2 * p + kek[d]);
  } else {
    upd(x, v, m + 1, e, d - 1, 2 * p + 1 - kek[d]);
  }
  a[p] = a[2 * p] + a[2 * p + 1];
}
long long query(int s, int e, int ps = 0, int pe = (1 << n) - 1, int d = n - 1,
                int p = 1) {
  if (e < ps || pe < s) return 0;
  if (s <= ps && pe <= e) return a[p];
  int pm = (ps + pe) / 2;
  return query(s, e, ps, pm, d - 1, 2 * p + kek[d]) +
         query(s, e, pm + 1, pe, d - 1, 2 * p + 1 - kek[d]);
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 0; i < (1 << n); i++) {
    int x;
    scanf("%d", &x);
    upd(i, x);
  }
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int x, v;
      scanf("%d %d", &x, &v);
      x--;
      upd(x, v);
    }
    if (t == 4) {
      int s, e;
      scanf("%d %d", &s, &e);
      printf("%lld\n", query(s - 1, e - 1));
    }
    if (t == 2) {
      int x;
      scanf("%d", &x);
      for (int i = 0; i < x; i++) kek[i] ^= 1;
    }
    if (t == 3) {
      int x;
      scanf("%d", &x);
      kek[x] ^= 1;
    }
  }
}
