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 maxn = 300005;
int n, q, len, rev[25], alrev[25];
long long sumv[maxn << 2];
inline int read() {
  int res = 0, f_f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f_f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    res = (res << 3) + (res << 1) + (ch - '0'), ch = getchar();
  return res * f_f;
}
inline void update(int o, int l, int r, int x, int v, int h, int cur) {
  if (l == r && l == x) {
    sumv[o] = v;
    return;
  }
  int mid = l + r >> 1, ls = o << 1, rs = o << 1 | 1;
  if (rev[h] ^ alrev[h] ^ cur) swap(ls, rs);
  if (x <= mid)
    update(ls, l, mid, x, v, h - 1, cur ^ alrev[h]);
  else
    update(rs, mid + 1, r, x, v, h - 1, cur ^ alrev[h]);
  sumv[o] = sumv[ls] + sumv[rs];
}
inline long long query(int o, int l, int r, int ql, int qr, int h, int cur) {
  if (ql <= l && r <= qr) {
    return sumv[o];
  }
  int mid = l + r >> 1, ls = o << 1, rs = o << 1 | 1;
  if (rev[h] ^ alrev[h] ^ cur) swap(ls, rs);
  if (qr <= mid)
    return query(ls, l, mid, ql, qr, h - 1, cur ^ alrev[h]);
  else if (ql > mid)
    return query(rs, mid + 1, r, ql, qr, h - 1, cur ^ alrev[h]);
  else
    return query(ls, l, mid, ql, qr, h - 1, cur ^ alrev[h]) +
           query(rs, mid + 1, r, ql, qr, h - 1, cur ^ alrev[h]);
}
int main() {
  n = read(), q = read(), len = 1 << n;
  for (int i = 1; i <= len; i++) {
    int x = read();
    update(1, 1, len, i, x, n - 1, 0);
  }
  while (q--) {
    int opt = read();
    if (opt == 1) {
      int x = read(), v = read();
      update(1, 1, len, x, v, n - 1, 0);
    } else if (opt == 2) {
      int x = read() - 1;
      alrev[x] ^= 1;
    } else if (opt == 3) {
      int x = read();
      rev[x] ^= 1;
    } else {
      int x = read(), y = read();
      printf("%lld\n", query(1, 1, len, x, y, n - 1, 0));
    }
  }
  return 0;
}
