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;
inline int read() {
  char c = 0;
  int x = 0;
  bool f = 0;
  while (!isdigit(c)) f = c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + c - 48, c = getchar();
  return f ? -x : x;
}
const int N = (1 << 18) + 100;
int a[N], n, q;
struct Segment_tree {
  int rev[N << 2];
  long long sum[N << 2];
  inline void pushup(int o) { sum[o] = sum[o << 1] + sum[o << 1 | 1]; }
  void build(int o, int l, int r) {
    if (l == r) {
      sum[o] = a[l];
      return;
    }
    register int mid = (l + r) >> 1;
    build(o << 1, l, mid);
    build(o << 1 | 1, mid + 1, r);
    pushup(o);
    return;
  }
  void update(int o, int l, int r, int dep, int p, int v) {
    if (l == r) {
      sum[o] = v;
      return;
    }
    register int mid = (l + r) >> 1;
    if (p <= mid)
      update((o << 1) | rev[dep], l, mid, dep - 1, p, v);
    else
      update(((o << 1) | rev[dep]) ^ 1, mid + 1, r, dep - 1, p, v);
    pushup(o);
    return;
  }
  long long query(int o, int l, int r, int dep, int p, int q) {
    if (p <= l && r <= q) return sum[o];
    register int mid = (l + r) >> 1;
    long long ans = 0;
    if (p <= mid) ans += query((o << 1) | rev[dep], l, mid, dep - 1, p, q);
    if (mid < q)
      ans += query(((o << 1) | rev[dep]) ^ 1, mid + 1, r, dep - 1, p, q);
    return ans;
  }
  inline void Reverse(int k) {
    for (int i = 0; i <= k; i++) rev[i] ^= 1;
  }
} T;
int main() {
  n = read();
  q = read();
  for (int i = 1; i <= (1 << n); i++) a[i] = read();
  T.build(1, 1, 1 << n);
  for (int t = 1, opt, l, r; t <= q; t++) {
    opt = read();
    if (opt == 1) {
      l = read();
      r = read();
      T.update(1, 1, 1 << n, n, l, r);
    } else if (opt == 2) {
      l = read();
      T.Reverse(l);
    } else if (opt == 3) {
      l = read();
      T.rev[l + 1] ^= 1;
    } else {
      l = read();
      r = read();
      printf("%lld\n", T.query(1, 1, 1 << n, n, l, r));
    }
  }
  return 0;
}
