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;
vector<long long> a, t;
long long LOL = 0;
void build(long long v, long long l, long long r) {
  if (l + 1 == r) {
    t[v] = a[l];
    return;
  }
  long long m = (l + r) / 2;
  build(v * 2 + 1, l, m);
  build(v * 2 + 2, m, r);
  t[v] = t[v * 2 + 1] + t[v * 2 + 2];
}
void update(long long v, long long l, long long r, long long i, long long x) {
  if (l + 1 == r) {
    t[v] = x;
    return;
  }
  long long m = (l + r) / 2;
  if (i < m)
    update(v * 2 + 1, l, m, i, x);
  else
    update(v * 2 + 2, m, r, i, x);
  t[v] = t[v * 2 + 1] + t[v * 2 + 2];
}
long long get(long long v, long long l, long long r, long long l1, long long r1,
              long long d) {
  if (l1 >= r || l >= r1) return 0;
  if (l1 <= l && r <= r1) return t[v];
  long long m = (l + r) / 2;
  bool is_flex = false;
  if (((1 << d) & LOL) > 0) is_flex = true;
  if (is_flex)
    return get(v * 2 + 1, m, r, l1, r1, d - 1) +
           get(v * 2 + 2, l, m, l1, r1, d - 1);
  else
    return get(v * 2 + 1, l, m, l1, r1, d - 1) +
           get(v * 2 + 2, m, r, l1, r1, d - 1);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long d, q;
  cin >> d >> q;
  long long n = (1 << d);
  a.resize(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  t.resize(4 * n);
  build(0, 0, n);
  while (q--) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long i, k;
      cin >> i >> k;
      i--;
      i ^= LOL;
      update(0, 0, n, i, k);
    }
    if (t == 2) {
      long long k;
      cin >> k;
      LOL ^= (1 << (k)) - 1;
    }
    if (t == 3) {
      long long k;
      cin >> k;
      LOL ^= (1 << (k));
    }
    if (t == 4) {
      long long l, r;
      cin >> l >> r;
      l--, r--;
      cout << get(0, 0, n, l, r + 1, d - 1) << '\n';
    }
  }
}
