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>
const bool debug = 0;
const long long MOD = 1000000007;
const double PI = acos(-1.0);
const double eps = 1e-9;
using namespace std;
bool flip[20];
int n;
int totalN;
long long seg[4 * 300100];
long long query(int a, int b, int l = 1, int r = totalN, int id = 1,
                int exp = n, int state = 0) {
  if (a > r || b < l) return 0;
  if (a <= l && b >= r) return seg[id];
  state ^= flip[exp];
  int m = (l + r) / 2;
  int ida = id * 2;
  int idb = id * 2 + 1;
  if (state) swap(ida, idb);
  return query(a, b, l, m, ida, exp - 1, state) +
         query(a, b, m + 1, r, idb, exp - 1, state);
}
void upd(int x, int pos, int l = 1, int r = totalN, int id = 1, int exp = n,
         int state = 0) {
  if (l == r) {
    seg[id] = x;
    return;
  }
  state ^= flip[exp];
  int m = (l + r) / 2;
  int ida = id * 2;
  int idb = id * 2 + 1;
  if (state) swap(ida, idb);
  if (pos <= m) {
    upd(x, pos, l, m, ida, exp - 1, state);
  } else
    upd(x, pos, m + 1, r, idb, exp - 1, state);
  seg[id] = seg[ida] + seg[idb];
}
int main() {
  int q;
  cin >> n >> q;
  totalN = 1 << n;
  assert(totalN < 300100);
  int x;
  for (int i = 1; i <= totalN; i++) {
    scanf("%d", &x);
    upd(x, i);
  }
  int a, b, c;
  for (int i = (0); i < (q); i++) {
    scanf("%d %d", &a, &b);
    if (a == 1) {
      scanf("%d", &c);
      upd(c, b);
    } else if (a == 2) {
      flip[b] ^= 1;
    } else if (a == 3) {
      flip[b] ^= 1;
      flip[b + 1] ^= 1;
    } else {
      scanf("%d", &c);
      printf("%lld\n", query(b, c));
    }
  }
  return 0;
}
