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 double eps = 1e-8;
inline int sign(double x) { return (x > eps) - (x < -eps); }
const int maxn = 3e5 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
long long sum[maxn << 2];
int lz[30];
void build(int l, int r, int rt) {
  if (l == r) {
    scanf("%d", &sum[rt]);
    return;
  }
  int m = l + r >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void upd(int p, int x, int k, int l, int r, int rt) {
  if (l == r) {
    sum[rt] = x;
    return;
  }
  int m = l + r >> 1;
  int lc = rt << 1, rc = rt << 1 | 1;
  if (lz[k]) swap(lc, rc);
  if (p <= m)
    upd(p, x, k - 1, l, m, lc);
  else
    upd(p, x, k - 1, m + 1, r, rc);
  sum[rt] = sum[lc] + sum[rc];
}
long long qry(int L, int R, int k, int l, int r, int rt) {
  if (L <= l && r <= R) return sum[rt];
  int m = l + r >> 1;
  long long res = 0;
  int lc = rt << 1, rc = rt << 1 | 1;
  if (lz[k]) swap(lc, rc);
  if (L <= m) res += qry(L, R, k - 1, l, m, lc);
  if (R > m) res += qry(L, R, k - 1, m + 1, r, rc);
  return res;
}
int main() {
  int m, q;
  cin >> m >> q;
  int n = 1 << m;
  build(1, n, 1);
  while (q--) {
    int ty, x;
    scanf("%d%d", &ty, &x);
    if (ty == 1) {
      int y;
      scanf("%d", &y);
      upd(x, y, m, 1, n, 1);
    } else if (ty == 4) {
      int y;
      scanf("%d", &y);
      printf("%lld\n", qry(x, y, m, 1, n, 1));
    } else if (ty == 2) {
      for (int i = x; i >= 0; --i) lz[i] ^= 1;
    } else
      lz[x + 1] ^= 1;
  }
  return 0;
}
