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>
#pragma GCC optimize("O3")
using namespace std;
const int maxn = (1 << 18) + 3;
int a[maxn];
int n, q;
void read() {
  cin >> n >> q;
  for (int i = 1; i <= (1 << n); i++) cin >> a[i];
}
struct node {
  long long sum;
  int l, r;
  bool rev, to_swap;
};
node tr[maxn * 4];
bool rev[20], to_swap[20];
void merge(int v) { tr[v].sum = tr[tr[v].l].sum + tr[tr[v].r].sum; }
void push(int v, int lvl) {
  if (lvl == n) return;
  tr[v].rev ^= rev[lvl];
  if (tr[v].rev) {
    swap(tr[tr[v].l], tr[tr[v].r]);
    tr[tr[v].l].rev ^= 1;
    tr[tr[v].r].rev ^= 1;
  }
  tr[v].rev = rev[lvl];
  tr[v].to_swap ^= to_swap[lvl];
  if (tr[v].to_swap) swap(tr[tr[v].l], tr[tr[v].r]);
  tr[v].to_swap = to_swap[lvl];
}
void build(int v, int l, int r) {
  if (l == r) {
    tr[v] = {a[l], 0, 0, false, false};
    return;
  }
  tr[v] = {0, v * 2, v * 2 + 1, false, false};
  int mid = (l + r) / 2;
  build(v * 2, l, mid);
  build(v * 2 + 1, mid + 1, r);
  merge(v);
}
void update(int v, int l, int r, int pos, int val, int lvl) {
  push(v, lvl);
  if (l == r) {
    tr[v].sum = val;
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(tr[v].l, l, mid, pos, val, lvl + 1);
  else
    update(tr[v].r, mid + 1, r, pos, val, lvl + 1);
  merge(v);
}
long long get(int v, int l, int r, int ll, int rr, int lvl) {
  push(v, lvl);
  if (l > rr || r < ll) return 0;
  if (l >= ll && r <= rr) return tr[v].sum;
  int mid = (l + r) / 2;
  return get(tr[v].l, l, mid, ll, rr, lvl + 1) +
         get(tr[v].r, mid + 1, r, ll, rr, lvl + 1);
}
void solve() {
  build(1, 1, (1 << n));
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int pos, val;
      cin >> pos >> val;
      update(1, 1, (1 << n), pos, val, 0);
    } else if (type == 2) {
      int k;
      cin >> k;
      rev[n - k] ^= 1;
    } else if (type == 3) {
      int k;
      cin >> k;
      to_swap[n - k - 1] ^= 1;
    } else {
      int l, r;
      cin >> l >> r;
      cout << get(1, 1, (1 << n), l, r, 0) << '\n';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  read();
  solve();
}
