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;
struct node {
  int l, r, ch[2];
  long long val;
};
node a[1000010];
int cnt, rot;
int val[1000010];
int n, q, pn;
bool rv[30];
inline void update(int k) { a[k].val = a[a[k].ch[0]].val + a[a[k].ch[1]].val; }
inline void build(int &k, int l, int r) {
  k = ++cnt;
  a[k].l = l, a[k].r = r;
  if (l == r) return a[k].val = val[l], void();
  int mid = l + r >> 1;
  build(a[k].ch[0], l, mid), build(a[k].ch[1], mid + 1, r);
  update(k);
}
inline long long query(int k, int l, int r, int ly) {
  if (a[k].l == l && a[k].r == r) return a[k].val;
  int mid = a[k].l + a[k].r >> 1;
  if (!rv[ly]) {
    if (r <= mid)
      return query(a[k].ch[0], l, r, ly - 1);
    else if (l > mid)
      return query(a[k].ch[1], l, r, ly - 1);
    else
      return query(a[k].ch[0], l, mid, ly - 1) +
             query(a[k].ch[1], mid + 1, r, ly - 1);
  } else {
    int haf = (a[k].r - a[k].l + 1) >> 1;
    if (r <= mid)
      return query(a[k].ch[1], l + haf, r + haf, ly - 1);
    else if (l > mid)
      return query(a[k].ch[0], l - haf, r - haf, ly - 1);
    else
      return query(a[k].ch[0], mid + 1 - haf, r - haf, ly - 1) +
             query(a[k].ch[1], l + haf, mid + haf, ly - 1);
  }
}
inline void change(int k, int tar, int val, int ly) {
  if (a[k].l == a[k].r) return a[k].val = val, void();
  int mid = a[k].l + a[k].r >> 1;
  if (!rv[ly]) {
    if (tar <= mid)
      change(a[k].ch[0], tar, val, ly - 1);
    else
      change(a[k].ch[1], tar, val, ly - 1);
  } else {
    int haf = (a[k].r - a[k].l + 1) >> 1;
    if (tar <= mid)
      change(a[k].ch[1], tar + haf, val, ly - 1);
    else
      change(a[k].ch[0], tar - haf, val, ly - 1);
  }
  update(k);
}
int type, l, r;
int main() {
  scanf("%d%d", &n, &q);
  pn = 1;
  for (int i = 1; i <= n; ++i) pn *= 2;
  for (int i = 1; i <= pn; ++i) scanf("%d", &val[i]);
  build(rot, 1, pn);
  for (int i = 1; i <= q; ++i) {
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d", &l, &r);
      change(rot, l, r, n);
    }
    if (type == 2) {
      scanf("%d", &l);
      for (int j = l; j >= 0; --j) rv[j] ^= 1;
    }
    if (type == 3) {
      scanf("%d", &l);
      rv[l + 1] ^= 1;
    }
    if (type == 4) {
      scanf("%d%d", &l, &r);
      printf("%lld\n", query(1, l, r, n));
    }
  }
  return 0;
}
