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 maxn = 1e6 + 10, inf = 1e9 + 7;
int root, tot, n, q, a[maxn], tag[maxn << 2], ls[maxn << 2], rs[maxn << 2];
long long sum[maxn << 2];
void build(int l, int r, int &p) {
  if (!p) p = ++tot;
  if (l == r) return sum[p] = a[l], void();
  int m = (l + r) >> 1;
  build(l, m, ls[p]), build(m + 1, r, rs[p]);
  sum[p] = sum[ls[p]] + sum[rs[p]];
}
void push_down(int l, int r, int p) {
  tag[ls[p]] ^= tag[p], tag[rs[p]] ^= tag[p];
  if (tag[p] & ((r - l + 1) >> 1)) swap(ls[p], rs[p]);
  tag[p] = 0;
}
void modify(int l, int r, int p, int x, int v) {
  push_down(l, r, p);
  if (l == r) return sum[p] = v, void();
  int m = (l + r) >> 1;
  if (x <= m)
    modify(l, m, ls[p], x, v);
  else
    modify(m + 1, r, rs[p], x, v);
  sum[p] = sum[ls[p]] + sum[rs[p]];
}
long long query(int l, int r, int p, int L, int R) {
  push_down(l, r, p);
  if (!p || l > R || r < L) return 0;
  if (L <= l && r <= R) return sum[p];
  long long ret = 0;
  int m = (l + r) >> 1;
  ret = query(l, m, ls[p], L, R) + query(m + 1, r, rs[p], L, R);
  sum[p] = sum[ls[p]] + sum[rs[p]];
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  n = 1 << n;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  build(1, n, root);
  while (q--) {
    int op, x, y;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d", &x, &y);
      modify(1, n, root, x, y);
    } else if (op == 2) {
      scanf("%d", &x);
      tag[root] ^= (1 << x) - 1;
    } else if (op == 3) {
      scanf("%d", &x);
      tag[root] ^= 1 << x;
    } else {
      scanf("%d%d", &x, &y);
      printf("%lld\n", query(1, n, root, x, y));
    }
  }
  return 0;
}
