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 = 20;
const long long mod = 1000000007;
vector<int> lazy(N, 0);
int n, m;
struct segment_tree {
  int B;
  vector<long long> sum;
  void init(int n) {
    B = 1;
    while (B < n) B *= 2;
    sum.resize(B * 2);
  }
  long long merge(long long a, long long b) { return a + b; }
  void build(int node, int l, int r, vector<long long> &a) {
    if (l == r - 1) {
      sum[node] = a[l];
      return;
    }
    int mid = (l + r) / 2;
    build(node * 2 + 1, l, mid, a);
    build(node * 2 + 2, mid, r, a);
    sum[node] = merge(sum[node * 2 + 1], sum[node * 2 + 2]);
  }
  void build(vector<long long> &a) { build(0, 0, B, a); }
  void upd(int node, int lvl, int pos, long long val) {
    if (lvl == 0) {
      sum[node] = val;
      return;
    }
    int lson = node * 2 + 1;
    int rson = node * 2 + 2;
    if (lazy[lvl]) swap(lson, rson);
    if (pos < (1 << (lvl - 1))) {
      upd(lson, lvl - 1, pos, val);
    } else {
      upd(rson, lvl - 1, pos - (1 << (lvl - 1)), val);
    }
    sum[node] = merge(sum[node * 2 + 1], sum[node * 2 + 2]);
  }
  void upd(int pos, int val) { upd(0, n, pos, val); }
  long long ask(int node, int lvl, int l, int r) {
    if (r <= 0 || l >= (1 << lvl)) {
      return 0;
    }
    if (l <= 0 && r >= (1 << lvl)) {
      return sum[node];
    }
    int lson = node * 2 + 1;
    int rson = node * 2 + 2;
    if (lazy[lvl]) {
      swap(lson, rson);
    }
    return merge(
        ask(lson, lvl - 1, l, r),
        ask(rson, lvl - 1, l - (1 << (lvl - 1)), r - (1 << (lvl - 1))));
  }
  long long ask(int l, int r) { return ask(0, n, l, r); }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  vector<long long> a(1 << n);
  for (long long &x : a) cin >> x;
  segment_tree st;
  st.init(1 << n);
  st.build(a);
  for (int i = 1; i <= m; i++) {
    int op;
    cin >> op;
    if (op == 1) {
      long long x, k;
      cin >> x >> k;
      st.upd(x - 1, k);
    } else if (op == 2) {
      long long k;
      cin >> k;
      for (int i = 0; i <= k; i++) {
        lazy[i] ^= 1;
      }
    } else if (op == 3) {
      long long k;
      cin >> k;
      lazy[k + 1] ^= 1;
    } else {
      long long l, r;
      cin >> l >> r;
      cout << st.ask(l - 1, r) << '\n';
    }
  }
  return 0;
}
