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 long long N = 1e6 + 228;
const long long oo = 1e18 + 228;
const long long K = 300;
mt19937_64 rnd(time(0));
int n;
int a[N];
bool d1[50], d2[50];
struct ST {
  vector<long long> t;
  void build(int v = 1, int l = 1, int r = n) {
    if (l == r)
      t[v] = a[l];
    else {
      int c = (l + r) / 2;
      build(v + v, l, c);
      build(v + v + 1, c + 1, r);
      t[v] = t[v + v] + t[v + v + 1];
    }
  }
  void uni() {
    t.resize(2 * N);
    build();
  }
  void upd(int x, int y, int v = 1, int l = 1, int r = n, int d = 1,
           int now = 0) {
    if (l == r) {
      t[v] = y;
      return;
    }
    int c = (l + r) / 2;
    now ^= d2[d];
    int now2 = now ^ d1[d];
    int v1 = v + v, v2 = v + v + 1;
    if (now2) swap(v1, v2);
    if (x <= c)
      upd(x, y, v1, l, c, d + 1, now);
    else
      upd(x, y, v2, c + 1, r, d + 1, now);
    t[v] = t[v + v] + t[v + v + 1];
  }
  long long get(int l, int r, int v = 1, int tl = 1, int tr = n, int d = 1,
                int now = 0) {
    if (l > r) return 0;
    if (tl == l && tr == r) return t[v];
    int c = (tl + tr) / 2;
    now ^= d2[d];
    int now2 = now ^ d1[d];
    int v1 = v + v, v2 = v + v + 1;
    if (now2) swap(v1, v2);
    return get(l, min(r, c), v1, tl, c, d + 1, now) +
           get(max(l, c + 1), r, v2, c + 1, tr, d + 1, now);
  }
} t;
void solve() {
  int q, x, y, l, r;
  cin >> n >> q;
  int w = n;
  n = (1 << n);
  for (int i = 1; i <= n; ++i) cin >> a[i];
  t.uni();
  for (int z = 1; z <= q; ++z) {
    int T;
    cin >> T;
    if (T == 1) {
      cin >> x >> y;
      t.upd(x, y);
    }
    if (T == 3) {
      cin >> x;
      d1[w - x] = 1 - d1[w - x];
    }
    if (T == 2) {
      cin >> x;
      d2[w - x + 1] = 1 - d2[w - x + 1];
    }
    if (T == 4) {
      cin >> l >> r;
      cout << ' ' << t.get(l, r) << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int T = 1;
  for (int z = 1; z <= T; ++z) solve();
  return 0;
}
