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 = (1 << 18) + 5;
long long a[maxn << 2], lz;
void build(int rt, int l, int r) {
  if (l == r) {
    cin >> a[rt];
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  a[rt] = a[rt << 1] + a[rt << 1 | 1];
}
void upd(int rt, int l, int r, int p, int v, int dep) {
  if (l == r) {
    a[rt] = v;
    return;
  }
  int mid = (l + r) >> 1;
  if (lz >> dep & 1) {
    if (p <= mid)
      upd(rt << 1 | 1, mid + 1, r, r - (mid - p + 1) + 1, v, dep - 1);
    else
      upd(rt << 1, l, mid, l + (p - mid) - 1, v, dep - 1);
  } else {
    if (p <= mid)
      upd(rt << 1, l, mid, p, v, dep - 1);
    else
      upd(rt << 1 | 1, mid + 1, r, p, v, dep - 1);
  }
  a[rt] = a[rt << 1] + a[rt << 1 | 1];
}
long long query(int rt, int l, int r, int ll, int rr, int dep) {
  if (ll == l && rr == r) {
    return a[rt];
  }
  int mid = (l + r) >> 1;
  long long res = 0;
  if (lz >> dep & 1) {
    if (rr <= mid)
      return query(rt << 1 | 1, mid + 1, r, ll + (r - mid), rr + (r - mid),
                   dep - 1);
    if (ll > mid)
      return query(rt << 1, l, mid, ll - (r - mid), rr - (r - mid), dep - 1);
    return query(rt << 1 | 1, mid + 1, r, r - (mid - ll + 1) + 1, r, dep - 1) +
           query(rt << 1, l, mid, l, l + (rr - mid) - 1, dep - 1);
  } else {
    if (rr <= mid) return query(rt << 1, l, mid, ll, rr, dep - 1);
    if (ll > mid) return query(rt << 1 | 1, mid + 1, r, ll, rr, dep - 1);
    return query(rt << 1, l, mid, ll, mid, dep - 1) +
           query(rt << 1 | 1, mid + 1, r, mid + 1, rr, dep - 1);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false), cout.tie(0), cin.tie(0);
  int n, q, m;
  cin >> n >> q;
  m = n - 1;
  n = 1 << n;
  build(1, 1, n);
  while (q--) {
    int op, l, r;
    cin >> op >> l;
    if (op == 1) {
      cin >> r;
      upd(1, 1, n, l, r, m);
    } else if (op == 2) {
      lz ^= (1 << l) - 1;
    } else if (op == 3) {
      lz ^= (1 << l);
    } else {
      cin >> r;
      cout << query(1, 1, n, l, r, m) << "\n";
    }
  }
  return 0;
}
