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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long maxn = (long long)3e6 + 5;
const long long mod = 998244353;
const long long inf = 0x3f3f3f3f;
long long T = 1;
bool rev[25];
long long n, q;
long long t[maxn];
void build(long long st, long long l, long long r) {
  if (l == r) {
    cin >> t[st];
    return;
  }
  long long mid = (l + r) >> 1;
  build(st << 1, l, mid);
  build(st << 1 | 1, mid + 1, r);
  t[st] = t[st << 1] + t[st << 1 | 1];
}
void update(long long st, long long l, long long r, long long dep, long long p,
            long long v) {
  if (l == r) {
    t[st] = v;
    return;
  }
  long long mid = (l + r) >> 1;
  if (p <= mid)
    update(st << 1 | rev[dep], l, mid, dep - 1, p, v);
  else
    update(st << 1 | rev[dep] ^ 1, mid + 1, r, dep - 1, p, v);
  t[st] = t[st << 1] + t[st << 1 | 1];
}
long long query(long long st, long long l, long long r, long long dep,
                long long ql, long long qr) {
  if (ql <= l && r <= qr) return t[st];
  long long mid = (l + r) >> 1;
  long long res = 0;
  if (ql <= mid) res += query(st << 1 | rev[dep], l, mid, dep - 1, ql, qr);
  if (qr > mid)
    res += query(st << 1 | rev[dep] ^ 1, mid + 1, r, dep - 1, ql, qr);
  return res;
}
void solve() {
  cin >> n >> q;
  build(1, 1, 1 << n);
  while (q--) {
    long long op;
    cin >> op;
    long long x, y;
    if (op == 1) {
      cin >> x >> y;
      update(1, 1, 1 << n, n, x, y);
    } else if (op == 2) {
      cin >> x;
      for (long long i = 1; i <= x; ++i) rev[i] ^= 1;
    } else if (op == 3) {
      cin >> x;
      rev[x + 1] ^= 1;
    } else {
      cin >> x >> y;
      cout << query(1, 1, 1 << n, n, x, y) << '\n';
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  while (T--) solve();
  return 0;
}
