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 = 1e6 + 9, Mod = 1e9 + 7;
int n, m, q, L, A[N], Xor;
long long SM[N];
long long Get(int l, int r) {
  long long rt = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) rt += SM[l++];
    if (r & 1) rt += SM[--r];
  }
  return rt;
}
inline void Modify(int i, int val) {
  A[i] = val;
  SM[i += n] = val;
  for (; i; i >>= 1) SM[i >> 1] = SM[i] + SM[i ^ 1];
}
inline long long GetF(int r) {
  long long tot = 0;
  while (r > 0) {
    int lg = __builtin_ctz(r);
    r -= r & -r;
    int le = r ^ Xor;
    le &= ((n - 1) - ((1 << lg) - 1));
    tot += Get(le, le + (1 << lg));
  }
  return tot;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> L >> q;
  n = 1 << L;
  for (int i = 0; i < n; i++) cin >> A[i], SM[i + n] = A[i];
  for (int i = n - 1; i; i--) SM[i] = (SM[i << 1] + SM[i << 1 ^ 1]);
  Xor = 0;
  for (; q; q--) {
    int tp;
    cin >> tp;
    if (tp == 1) {
      int i, val;
      cin >> i >> val;
      i = (i - 1) ^ Xor;
      Modify(i, val);
      continue;
    } else if (tp == 2) {
      int k;
      cin >> k;
      Xor ^= (1 << k) - 1;
      continue;
    } else if (tp == 3) {
      int k;
      cin >> k;
      Xor ^= (1 << k);
    } else {
      int l, r;
      cin >> l >> r;
      long long tot = GetF(r) - GetF(l - 1);
      printf("%lld\n", tot);
    }
  }
  return 0;
}
