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;
long long a[300005], seg[1050005], n, q, flag[25];
void build(long long l, long long r, long long i) {
  if (l == r) {
    seg[i] = a[l];
    return;
  }
  long long m = (l + r) / 2;
  build(l, m, i * 2);
  build(m + 1, r, i * 2 + 1);
  seg[i] = seg[i * 2] + seg[i * 2 + 1];
  return;
}
long long query(long long l, long long r, long long i, long long lq,
                long long rq, long long d) {
  if (r < lq || l > rq) return 0;
  if (lq <= l && r <= rq) return seg[i];
  long long m = (l + r) / 2;
  if (flag[d])
    return query(l, m, i * 2 + 1, lq, rq, d - 1) +
           query(m + 1, r, i * 2, lq, rq, d - 1);
  return query(l, m, i * 2, lq, rq, d - 1) +
         query(m + 1, r, i * 2 + 1, lq, rq, d - 1);
}
void updoot(long long l, long long r, long long i, long long k, long long x) {
  seg[i] += k;
  if (l == r) return;
  long long m = (l + r) / 2;
  if (x <= m)
    updoot(l, m, i * 2, k, x);
  else
    updoot(m + 1, r, i * 2 + 1, k, x);
}
void winterDaydreams(long long x, long long k) {
  for (long long i = 0; i < n; i++) {
    if (flag[i]) x = ((x - 1) ^ (1 << i)) + 1;
  }
  k -= a[x];
  a[x] += k;
  updoot(1, 1 << n, 1, k, x);
  return;
}
void littleRussian(long long k) {
  for (long long i = 0; i < k; i++) flag[i] ^= 1;
  return;
}
void polish(long long k) {
  flag[k] ^= 1;
  return;
}
void fate(long long l, long long r) {
  cout << query(1, 1 << n, 1, l, r, n - 1) << endl;
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> q;
  for (long long i = 1; i <= 1 << n; i++) {
    cin >> a[i];
  }
  build(1, 1 << n, 1);
  for (long long i = 1; i <= q; i++) {
    long long tmp, tmp1, tmp2;
    cin >> tmp;
    if (tmp == 1) cin >> tmp1 >> tmp2, winterDaydreams(tmp1, tmp2);
    if (tmp == 2) cin >> tmp1, littleRussian(tmp1);
    if (tmp == 3) cin >> tmp1, polish(tmp1);
    if (tmp == 4) cin >> tmp1 >> tmp2, fate(tmp1, tmp2);
  }
  return 0;
}
