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 long long N = 1000005;
long long sum[N << 2], n, m, q, a[N], num[25], p[25];
void build(long long x, long long l, long long r) {
  if (l == r) {
    sum[x] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(x << 1, l, mid);
  build(x << 1 | 1, mid + 1, r);
  sum[x] = sum[x << 1] + sum[x << 1 | 1];
}
void update(long long x, long long l, long long r, long long wh, long long k,
            long long dep) {
  if (l == r) {
    sum[x] = k;
    return;
  }
  long long mid = (l + r) >> 1;
  if (num[dep])
    if (wh <= mid)
      wh += (p[dep] >> 1);
    else
      wh -= (p[dep] >> 1);
  if (wh <= mid)
    update(x << 1, l, mid, wh, k, dep + 1);
  else
    update(x << 1 | 1, mid + 1, r, wh, k, dep + 1);
  sum[x] = sum[x << 1] + sum[x << 1 | 1];
}
long long qurey(long long x, long long l, long long r, long long L, long long R,
                long long dep) {
  if (l > R || r < L) return 0;
  if (L <= l && r <= R) return sum[x];
  long long ans = 0, mid = (l + r) >> 1;
  if (num[dep])
    ans += qurey(x << 1, l, mid, L - (p[dep] >> 1), R - (p[dep] >> 1), dep + 1);
  else
    ans += qurey(x << 1, l, mid, L, R, dep + 1);
  if (num[dep])
    ans += qurey(x << 1 | 1, mid + 1, r, L + (p[dep] >> 1), R + (p[dep] >> 1),
                 dep + 1);
  else
    ans += qurey(x << 1 | 1, mid + 1, r, L, R, dep + 1);
  return ans;
}
signed main() {
  scanf("%lld%lld", &m, &q);
  n = (1 << m);
  p[0] = n;
  for (long long i = 1; i <= m; i++) p[i] = (p[i - 1] >> 1);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    sum[i] = sum[i - 1] + a[i];
  }
  build(1, 1, n);
  while (q--) {
    long long opt, x, y;
    scanf("%lld", &opt);
    if (opt == 1) scanf("%lld%lld", &x, &y), update(1, 1, n, x, y, 0);
    if (opt == 2) {
      scanf("%lld", &x);
      for (long long i = m - x; i <= 20; i++) num[i] ^= 1;
    }
    if (opt == 3) {
      scanf("%lld", &x);
      num[m - x - 1] ^= 1;
    }
    if (opt == 4)
      scanf("%lld%lld", &x, &y), printf("%lld\n", qurey(1, 1, n, x, y, 0));
  }
  return 0;
}
