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 = 1e6;
int L[N], R[N], sp[N], levsp[30], lev[N];
long long val[N], a[N];
void lazyUpdate(int node) {
  if (!L[node]) return;
  if (sp[node] ^ levsp[lev[node]]) {
    swap(L[node], R[node]);
    sp[node] ^= 1;
  }
}
long long build(int node, int l, int r, int lv) {
  if (l == r) return val[node] = a[l];
  int mid = l + r >> 1;
  lev[node] = lv;
  L[node] = node << 1, R[node] = node << 1 | 1;
  val[node] =
      build(L[node], l, mid, lv - 1) + build(R[node], mid + 1, r, lv - 1);
  return val[node];
}
long long update(int node, int l, int r, int pos, int k) {
  lazyUpdate(node);
  if (r < pos || l > pos) return val[node];
  if (l == r) return val[node] = k;
  int mid = l + r >> 1;
  return val[node] = update(L[node], l, mid, pos, k) +
                     update(R[node], mid + 1, r, pos, k);
}
long long query(int node, int l, int r, int pos) {
  lazyUpdate(node);
  if (l > pos) return 0;
  if (r <= pos) return val[node];
  int mid = l + r >> 1;
  return query(L[node], l, mid, pos) + query(R[node], mid + 1, r, pos);
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  int nn = 1 << n;
  for (int i = 1; i <= nn; i++) scanf("%lld", &a[i]);
  build(1, 1, nn, n);
  while (q--) {
    int t, x, k;
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d", &x, &k);
      update(1, 1, nn, x, k);
    } else if (t == 2) {
      scanf("%d", &k);
      for (int i = 0; i <= k; i++) levsp[i] ^= 1;
    } else if (t == 3) {
      scanf("%d", &k);
      levsp[k + 1] ^= 1;
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%lld\n", query(1, 1, nn, r) - query(1, 1, nn, l - 1));
    }
  }
  return 0;
}
