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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < '0' || c > '9'; c = getchar())
    if (!(c ^ 45)) f = 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
}
long long K, Q, n, rt, a[300005], rev[25], t[1200005];
inline void pushup(long long x) { t[x] = t[x << 1] + t[x << 1 | 1]; }
inline void build(long long x, long long l, long long r) {
  if (l == r) return (void)(t[x] = a[l]);
  build(x << 1, l, (l + r) >> 1), build(x << 1 | 1, ((l + r) >> 1) + 1, r),
      pushup(x);
}
inline long long query(long long x, long long l, long long r, long long dl,
                       long long dr, long long dep) {
  if (dl > r || l > dr)
    return 0;
  else if (dl <= l && r <= dr)
    return t[x];
  return query(x << 1 | rev[dep], l, (l + r) >> 1, dl, dr, dep + 1) +
         query(x << 1 | !rev[dep], ((l + r) >> 1) + 1, r, dl, dr, dep + 1);
}
inline void motif(long long x, long long l, long long r, long long dw,
                  long long dc, long long dep) {
  if (l == r) return (void)(t[x] = dc);
  if (dw <= ((l + r) >> 1))
    motif(x << 1 | rev[dep], l, (l + r) >> 1, dw, dc, dep + 1), pushup(x);
  else
    motif(x << 1 | !rev[dep], ((l + r) >> 1) + 1, r, dw, dc, dep + 1),
        pushup(x);
}
signed main() {
  read(K), read(Q), n = 1 << K;
  for (long long i = 1; i <= n; i++) read(a[i]);
  build(1, 1, n);
  for (long long fg, x, y; Q--;) {
    read(fg), read(x);
    if (fg == 1) read(y), motif(1, 1, n, x, y, 1);
    if (fg == 2)
      for (long long i = K + 1; i >= K - x + 1; i--) rev[i] ^= 1;
    if (fg == 3) rev[K - x] ^= 1;
    if (fg == 4) read(y), printf("%lld\n", query(1, 1, n, x, y, 1));
  }
  return 0;
}
