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 N = (1 << 18) + 1;
int n;
long long fen[N];
void update(int x, int v) {
  for (; x <= 1 << n; x += x & -x) fen[x] += v;
}
long long query(int x) {
  long long res = 0;
  for (; x; x -= x & -x) res += fen[x];
  return res;
}
long long query(int l, int r) { return query(r) - query(l - 1); }
int hbit(int x) {
  x |= x >> 1;
  x |= x >> 2;
  x |= x >> 4;
  x |= x >> 8;
  x |= x >> 16;
  return x ^ x >> 1;
}
long long get(int l, int r, int x) {
  long long res = 0;
  int len = r - l + 1, t = min(l & -l, hbit(len));
  if (!t) t = hbit(len);
  while (len) {
    int tmp = (l ^ x) & ~(t - 1);
    res += query(tmp + 1, tmp + t);
    len -= t, l += t, t = min(l & -l, hbit(len));
    if (!t) t = hbit(len);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int q;
  cin >> n >> q;
  for (int i = 0; i < 1 << n; i++) {
    int x;
    cin >> x;
    update(i + 1, x);
  }
  int cur = 0;
  for (int i = 0; i < q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int x, k;
      cin >> x >> k;
      x = (x - 1 ^ cur) + 1;
      update(x, k - query(x, x));
    } else if (t == 2) {
      int k;
      cin >> k;
      cur ^= (1 << k) - 1;
    } else if (t == 3) {
      int k;
      cin >> k;
      cur ^= 1 << k;
    } else if (t == 4) {
      int l, r;
      cin >> l >> r;
      cout << get(l - 1, r - 1, cur) << '\n';
    }
  }
}
