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;
const int N = (1 << 18) + 10;
struct node {
  int son[2];
  int tag;
  long long sum;
} f[N * 4];
int a[N];
int n, m;
void upd(int s) { f[s].sum = f[f[s].son[0]].sum + f[f[s].son[1]].sum; }
void build(int l, int r, int s) {
  if (l == r) {
    f[s].sum = a[l];
    return;
  }
  f[s].son[0] = s + s;
  f[s].son[1] = s + s + 1;
  build(l, (l + r) / 2, s + s);
  build((l + r) / 2 + 1, r, s + s + 1);
  upd(s);
}
void down(int l, int r, int s) {
  if (f[s].tag) {
    if (f[s].tag & 1) swap(f[s].son[0], f[s].son[1]);
    if (l != r)
      f[f[s].son[0]].tag ^= (f[s].tag >> 1),
          f[f[s].son[1]].tag ^= (f[s].tag >> 1);
    f[s].tag = 0;
  }
}
void change(int l, int r, int s, int ll, int v) {
  down(l, r, s);
  if (l == r) {
    f[s].sum = v;
    return;
  }
  if ((l + r) / 2 >= ll)
    change(l, (l + r) / 2, f[s].son[0], ll, v);
  else
    change((l + r) / 2 + 1, r, f[s].son[1], ll, v);
  upd(s);
}
long long get(int l, int r, int s, int ll, int rr) {
  down(l, r, s);
  if (r < ll || rr < l) return 0;
  if (ll <= l && r <= rr) {
    return f[s].sum;
  }
  return get(l, (l + r) / 2, f[s].son[0], ll, rr) +
         get((l + r) / 2 + 1, r, f[s].son[1], ll, rr);
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= (1 << n); i++) scanf("%d", &a[i]);
  build(1, (1 << n), 1);
  while (m--) {
    int sig, x, y;
    scanf("%d %d", &sig, &x);
    switch (sig) {
      case 1:
        scanf("%d", &y);
        change(1, (1 << n), 1, x, y);
        break;
      case 2:
        f[1].tag ^= (((1 << n) - 1) - ((1 << (n - x)) - 1));
        break;
      case 3:
        f[1].tag ^= (1 << (n - x - 1));
        break;
      case 4:
        scanf("%d", &y);
        printf("%lld\n", get(1, (1 << n), 1, x, y));
        break;
    }
  }
  return 0;
}
