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 MOD = (int)7 + (int)1e9;
const int N = 4000005;
const double PI = 4 * atan(1);
const double eps = 1e-10;
const long long oo = 1e10;
const int K = 26;
long long tree[N];
bool rev[20];
int n;
int q;
long long a[N];
int ty;
int x, l, r;
long long k;
void build(int l, int r, int pos) {
  if (l == r) {
    tree[pos] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, 2 * pos + 1);
  build(mid + 1, r, 2 * pos + 2);
  tree[pos] = tree[2 * pos + 1] + tree[2 * pos + 2];
}
void update(int l, int r, int pos, int id, long long value) {
  if (l > id) return;
  if (r < id) return;
  if (l == r && l == id) {
    tree[pos] = value;
    return;
  }
  int mid = (l + r) / 2;
  int pos1 = 2 * pos + 1;
  int pos2 = 2 * pos + 2;
  int l2 = log2(r - l + 1);
  if (rev[l2]) swap(pos1, pos2);
  update(l, mid, pos1, id, value);
  update(mid + 1, r, pos2, id, value);
  tree[pos] = tree[pos1] + tree[pos2];
}
long long getval(int l, int r, int pos, int a, int b) {
  if (l > b || r < a) return 0ll;
  if (l >= a && r <= b) {
    return tree[pos];
  }
  int mid = (l + r) / 2;
  int pos1 = 2 * pos + 1;
  int pos2 = 2 * pos + 2;
  int l2 = log2(r - l + 1);
  if (rev[l2]) swap(pos1, pos2);
  return getval(l, mid, pos1, a, b) + getval(mid + 1, r, pos2, a, b);
}
int main() {
  srand(time(NULL));
  scanf("%d %d", &n, &q);
  for (int i = 0; i < (1 << n); i++) {
    scanf("%d", a + i);
  }
  build(0, (1 << n) - 1, 0);
  while (q--) {
    scanf("%d", &ty);
    if (ty == 1) {
      scanf("%d %lld", &x, &k);
      update(0, (1 << n) - 1, 0, x - 1, k);
    } else if (ty == 2) {
      scanf("%lld", &k);
      for (int i = 0; i <= k; i++) {
        rev[i] ^= 1;
      }
    } else if (ty == 3) {
      scanf("%lld", &k);
      rev[k + 1] ^= 1;
    } else {
      scanf("%d %d", &l, &r);
      long long ans = getval(0, (1 << n) - 1, 0, l - 1, r - 1);
      printf("%lld\n", ans);
    }
  }
  return 0;
}
