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;
long long n, q, rot[20], a[600000], cs[300001], p;
inline void read(long long &res) {
  res = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') {
    res = (res << 1) + (res << 3) + c - 48;
    c = getchar();
  }
}
void build(long long l, long long r, long long rt) {
  if (l == r) {
    a[rt] = cs[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(l, mid, (rt << 1));
  build(mid + 1, r, ((rt << 1) ^ 1));
  a[rt] = a[(rt << 1)] + a[((rt << 1) ^ 1)];
}
inline void modify(long long mb, long long L, long long R, long long rt,
                   long long fz, long long ccs) {
  if (L == mb && R == mb) {
    a[rt] = fz;
    return;
  }
  long long mid = (L + R) >> 1;
  if (mid >= mb)
    modify(mb, L, mid, (rt << 1) ^ rot[ccs], fz, ccs - 1);
  else
    modify(mb, mid + 1, R, ((rt << 1) ^ 1) ^ rot[ccs], fz, ccs - 1);
  a[rt] = a[(rt << 1)] + a[((rt << 1) ^ 1)];
}
long long query(long long l, long long r, long long L, long long R,
                long long rt, long long ccs) {
  if (L >= l && r >= R) return a[rt];
  long long mid = (L + R) >> 1, ll = 0, rr = 0;
  if (mid >= l) ll = query(l, r, L, mid, (rt << 1) ^ rot[ccs], ccs - 1);
  if (mid < r)
    rr = query(l, r, mid + 1, R, ((rt << 1) ^ 1) ^ rot[ccs], ccs - 1);
  return ll + rr;
}
signed main() {
  read(n);
  read(p);
  long long xx = pow(2, n);
  for (long long i = 1; i <= xx; i++) read(cs[i]);
  build(1, xx, 1);
  long long op, ll, rr;
  for (long long i = 1; i <= p; i++) {
    read(op);
    if (op == 1) {
      read(ll);
      read(rr);
      modify(ll, 1, xx, 1, rr, n);
    }
    if (op == 2) {
      read(ll);
      for (long long i = 0; i <= ll; i++) rot[i] ^= 1;
    }
    if (op == 3) {
      read(ll);
      rot[ll + 1] ^= 1;
    }
    if (op == 4) {
      read(ll);
      read(rr);
      printf("%lld\n", query(ll, rr, 1, xx, 1, n));
    }
  }
  return 0;
}
