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>
constexpr auto Inf = 0X3F3F3F3F;
using namespace std;
namespace IO {
inline long long read() {
  long long o = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c > '/' && c < ':') {
    o = o * 10 + c - '0';
    c = getchar();
  }
  return o * f;
}
inline char recd() {
  char o;
  while ((o = getchar()) < '0' || o > '9')
    ;
  return o;
}
inline double reod() {
  double o = read(), f = 1;
  char c;
  while ((c = getchar()) > '/' && c < ':') o += (c - '0') / (f *= 10);
  return o;
}
}  // namespace IO
using namespace IO;
const int SIZE = 2E5 + 7, Mod = 1E9 + 7;
struct Nodde {
  int L, R;
  long long S;
} Node[1 << 19 | 1];
int swp[23];
void pushup(int now) { Node[now].S = Node[now << 1].S + Node[now << 1 | 1].S; }
void BUILD(int now, int L, int R) {
  Node[now].L = L, Node[now].R = R;
  if (L == R) {
    Node[now].S = read();
    return;
  }
  int M = (L + R) >> 1;
  BUILD(now << 1, L, M), BUILD(now << 1 | 1, M + 1, R);
  pushup(now);
}
void Upd(int now, int L, int R, int pos, int w, int dep) {
  if (L == R) {
    Node[now].S = w;
    return;
  }
  int M = (L + R) >> 1;
  if (M < pos)
    Upd((now << 1 | 1) - swp[dep], M + 1, R, pos, w, dep - 1);
  else
    Upd((now << 1) + swp[dep], L, M, pos, w, dep - 1);
  pushup(now);
}
long long Ask(int now, int L, int R, int aL, int aR, int dep) {
  if (L >= aL && R <= aR) return Node[now].S;
  int M = (L + R) >> 1;
  if (M < aL)
    return Ask((now << 1 | 1) - swp[dep], M + 1, R, aL, aR, dep - 1);
  else if (M >= aR)
    return Ask((now << 1) + swp[dep], L, M, aL, aR, dep - 1);
  else
    return Ask((now << 1) + swp[dep], L, M, aL, M, dep - 1) +
           Ask((now << 1 | 1) - swp[dep], M + 1, R, M + 1, aR, dep - 1);
}
int main() {
  int N = read(), M = read();
  BUILD(1, 1, 1 << N);
  int o, L, R;
  while (M--) {
    o = read(), L = read();
    if (o == 1)
      R = read(), Upd(1, 1, 1 << N, L, R, N);
    else if (o == 2)
      for (int pos = 0; pos <= L; pos++) swp[pos] ^= 1;
    else if (o == 3)
      swp[L + 1] ^= 1;
    else
      R = read(), printf("%lld\n", Ask(1, 1, 1 << N, L, R, N));
  }
}
