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 {
  long long sum;
  int dep;
};
int N, Q;
Node seg[1000000];
long long arr[1000000];
bool rev[20];
void build(int l, int r, int idx) {
  seg[idx].dep = 31 - __builtin_clz(r - l + 1);
  if (l == r) {
    seg[idx].sum = arr[l];
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, 2 * idx);
  build(mid + 1, r, 2 * idx + 1);
  seg[idx].sum = seg[2 * idx].sum + seg[2 * idx + 1].sum;
}
void upd(int p, int v, int idx) {
  if (seg[idx].dep == 0) {
    seg[idx].sum = v;
    return;
  }
  int half = (1 << (seg[idx].dep - 1));
  if (p >= half) {
    upd(p - half, v, 2 * idx + (1 ^ rev[seg[idx].dep]));
  } else {
    upd(p, v, 2 * idx + (rev[seg[idx].dep]));
  }
  seg[idx].sum = seg[2 * idx].sum + seg[2 * idx + 1].sum;
}
long long query(int l, int r, int idx) {
  if (l + (1 << seg[idx].dep) - 1 == r) {
    return seg[idx].sum;
  }
  int half = (1 << (seg[idx].dep - 1));
  if (r < half) {
    return query(l, r, 2 * idx + rev[seg[idx].dep]);
  } else if (l >= half) {
    return query(l - half, r - half, 2 * idx + (1 ^ rev[seg[idx].dep]));
  } else {
    return query(l, half - 1, 2 * idx + rev[seg[idx].dep]) +
           query(0, r - half, 2 * idx + (1 ^ rev[seg[idx].dep]));
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> Q;
  for (int i = 0; i < 1 << N; i++) {
    cin >> arr[i];
  }
  build(0, (1 << N) - 1, 1);
  while (Q--) {
    int c;
    cin >> c;
    if (c == 1) {
      int p, x;
      cin >> p >> x;
      p--;
      arr[p] = x;
      upd(p, x, 1);
    } else if (c == 2) {
      int k;
      cin >> k;
      for (int i = k; i >= 0; i--) {
        rev[i] ^= 1;
      }
    } else if (c == 3) {
      int k;
      cin >> k;
      rev[1 + k] ^= 1;
    } else {
      int l, r;
      cin >> l >> r;
      l--, r--;
      cout << query(l, r, 1) << "\n";
    }
  }
}
