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;
struct Segment {
  long long sum;
};
Segment seg[8 * 500005];
long long n, q, a[500005], rev[500005];
long long lson(long long p) { return 2 * p; }
long long rson(long long p) { return 2 * p + 1; }
void push_up(long long p) { seg[p].sum = seg[lson(p)].sum + seg[rson(p)].sum; }
void build(long long l, long long r, long long p) {
  if (l == r) {
    seg[p].sum = a[l];
    return;
  }
  long long mid = (l + r) / 2;
  build(l, mid, lson(p));
  build(mid + 1, r, rson(p));
  push_up(p);
}
void update(long long l, long long r, long long p, long long dep, long long x,
            long long k) {
  if (l == r) {
    seg[p].sum = k;
    return;
  }
  long long mid = (l + r) / 2;
  if (x <= mid)
    update(l, mid, lson(p) ^ rev[dep], dep - 1, x, k);
  else
    update(mid + 1, r, rson(p) ^ rev[dep], dep - 1, x, k);
  push_up(p);
}
long long query(long long l, long long r, long long p, long long dep,
                long long nl, long long nr) {
  long long res = 0;
  if (l >= nl && r <= nr) {
    return seg[p].sum;
  }
  long long mid = (l + r) / 2;
  if (nl <= mid) res += query(l, mid, lson(p) ^ rev[dep], dep - 1, nl, nr);
  if (nr > mid) res += query(mid + 1, r, rson(p) ^ rev[dep], dep - 1, nl, nr);
  push_up(p);
  return res;
}
signed main(void) {
  cin >> n >> q;
  long long d = n;
  n = (1 << n);
  for (long long i = 1; i <= n; i++) cin >> a[i];
  build(1, n, 1);
  while (q--) {
    long long opt;
    cin >> opt;
    if (opt == 1) {
      long long x, k;
      cin >> x >> k;
      update(1, n, 1, d, x, k);
    }
    if (opt == 2) {
      long long k;
      cin >> k;
      for (long long i = 0; i <= k; i++) rev[i] ^= 1;
    }
    if (opt == 3) {
      long long k;
      cin >> k;
      rev[k + 1] ^= 1;
    }
    if (opt == 4) {
      long long l, r;
      cin >> l >> r;
      cout << query(1, n, 1, d, l, r) << '\n';
    }
  }
  return 0;
}
