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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const double PI = acos(-1.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 2e6 + 100;
long long lpt[N], rpt[N];
long long sum[N];
long long lazy[N];
long long n, q;
void propagate(long long node, long long l, long long r, long long lvl) {
  if (lazy[node] & (1LL << lvl)) {
    swap(lpt[node], rpt[node]);
    lazy[node] ^= (1 << lvl);
  }
  if (l != r) {
    lazy[lpt[node]] ^= lazy[node];
    lazy[rpt[node]] ^= lazy[node];
  }
  lazy[node] = 0;
}
void read(long long node, long long l, long long r) {
  if (l == r) {
    cin >> sum[node];
    return;
  } else {
    read(lpt[node], l, (l + r) / 2);
    read(rpt[node], 1 + (l + r) / 2, r);
    sum[node] = sum[lpt[node]] + sum[rpt[node]];
  }
}
void update(long long node, long long l, long long r, long long lvl,
            long long msk) {
  lazy[node] ^= msk;
  return;
}
void replace(long long node, long long l, long long r, long long lvl,
             long long ind, long long val) {
  if (l > ind || r < ind) return;
  propagate(node, l, r, lvl);
  if (l == r) {
    sum[node] = val;
    return;
  }
  long long mid = (l + r) / 2;
  replace(lpt[node], l, mid, lvl - 1, ind, val);
  replace(rpt[node], mid + 1, r, lvl - 1, ind, val);
  sum[node] = sum[lpt[node]] + sum[rpt[node]];
  return;
}
long long query(long long node, long long l, long long r, long long lvl,
                long long ql, long long qr) {
  if (ql > r || l > qr) {
    return 0;
  }
  if (ql <= l && r <= qr) return sum[node];
  propagate(node, l, r, lvl);
  long long mid = (l + r) / 2;
  return query(lpt[node], l, mid, lvl - 1, ql, qr) +
         query(rpt[node], mid + 1, r, lvl - 1, ql, qr);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (long long i = 0; i < N; i++) {
    lpt[i] = 2 * i + 1;
    rpt[i] = 2 * i + 2;
  }
  read(0, 1, (1 << n));
  while (q--) {
    long long type;
    cin >> type;
    if (type == 1) {
      long long x, k;
      cin >> x >> k;
      replace(0, 1, 1 << n, n, x, k);
    } else if (type == 2) {
      long long k;
      cin >> k;
      update(0, 1, 1 << n, n, (1 << (k + 1)) - 1);
    } else if (type == 3) {
      long long k;
      cin >> k;
      update(0, 1, 1 << n, n, (1 << (k + 1)));
    } else {
      long long l, r;
      cin >> l >> r;
      cout << query(0, 1, 1 << n, n, l, r) << "\n";
    }
  }
}
