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 a[300007];
struct node {
  long long sum;
  int len;
  node *pl, *pr;
  int swap_lazy, rev_lazy;
  node(int _len, int _val) {
    this->len = _len;
    this->sum = _val;
    this->pl = this->pr = NULL;
    this->swap_lazy = this->rev_lazy = 0;
  }
};
void unite(node *aux) {
  aux->sum = 0;
  if (aux->pl != NULL) {
    aux->sum += aux->pl->sum;
  }
  if (aux->pr != NULL) {
    aux->sum += aux->pr->sum;
  }
}
void push_lazy(node *aux) {
  int hh1 = aux->swap_lazy & 1;
  int hh2 = aux->rev_lazy & 1;
  if (hh1 > 0) {
    swap(aux->pl, aux->pr);
  }
  if (hh2 > 0) {
    swap(aux->pl, aux->pr);
    if (aux->pl != NULL) {
      aux->pl->rev_lazy ^= 1;
    }
    if (aux->pr != NULL) {
      aux->pr->rev_lazy ^= 1;
    }
  }
  if (aux->pl != NULL) {
    aux->pl->swap_lazy ^= (aux->swap_lazy >> 1);
    aux->pl->rev_lazy ^= (aux->rev_lazy >> 1);
  }
  if (aux->pr != NULL) {
    aux->pr->swap_lazy ^= (aux->swap_lazy >> 1);
    aux->pr->rev_lazy ^= (aux->rev_lazy >> 1);
  }
  aux->swap_lazy = aux->rev_lazy = 0;
}
void init_tree(node *(&w), int l, int r) {
  if (l == r) {
    w = new node(r - l + 1, a[l]);
    return;
  }
  w = new node(r - l + 1, 0);
  int mid = (l + r) / 2;
  init_tree(w->pl, l, mid);
  init_tree(w->pr, mid + 1, r);
  unite(w);
}
void update_pos(node *w, int pos, int val, int coef) {
  push_lazy(w);
  if (coef + w->len < pos || pos < coef + 1) {
    return;
  }
  if (w->len == 1) {
    w->sum = val;
    return;
  }
  update_pos(w->pl, pos, val, coef);
  update_pos(w->pr, pos, val, coef + w->pl->len);
  unite(w);
}
long long query(node *w, int st, int en, int coef) {
  push_lazy(w);
  if (coef + w->len < st || en < coef + 1) {
    return 0;
  }
  if (st <= coef + 1 && coef + w->len <= en) {
    return w->sum;
  }
  return query(w->pl, st, en, coef) + query(w->pr, st, en, coef + w->pl->len);
}
node *root;
void input() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= (1 << n); ++i) {
    scanf("%d", &a[i]);
  }
  init_tree(root, 1, (1 << n));
}
void solve() {
  int type, x, y;
  while (q--) {
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d", &x, &y);
      update_pos(root, x, y, 0);
    } else if (type == 2) {
      scanf("%d", &x);
      root->rev_lazy ^= (1 << (n - x));
    } else if (type == 3) {
      scanf("%d", &x);
      root->swap_lazy ^= (1 << (n - x - 1));
    } else {
      scanf("%d%d", &x, &y);
      printf("%lld\n", query(root, x, y, 0));
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}
