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>
const int maxn = 3e5 + 1;
inline long long max(long long a, long long b) { return a > b ? a : b; }
inline long long min(long long a, long long b) { return a < b ? a : b; }
long long rev[19], n, m, opt, k, x, l, r;
struct Segment_Tree {
  long long val[maxn << 2];
  inline void pushup(long long rt) {
    val[rt] = val[rt << 1] + val[rt << 1 | 1];
  }
  inline void build(long long rt, long long l, long long r) {
    if (l == r) return (void)(scanf("%lld", &val[rt]));
    build(rt << 1, l, ((l + r) >> 1));
    build(rt << 1 | 1, ((l + r) >> 1) + 1, r);
    pushup(rt);
  }
  inline void update(long long rt, long long l, long long r, long long k,
                     long long ch, long long dep) {
    if (l == r) return (void)(val[rt] = ch);
    if (k > ((l + r) >> 1)) {
      if (rev[dep])
        update(rt << 1, ((l + r) >> 1) + 1, r, k, ch, dep - 1);
      else
        update(rt << 1 | 1, ((l + r) >> 1) + 1, r, k, ch, dep - 1);
    } else {
      if (rev[dep])
        update(rt << 1 | 1, l, ((l + r) >> 1), k, ch, dep - 1);
      else
        update(rt << 1, l, ((l + r) >> 1), k, ch, dep - 1);
    }
    pushup(rt);
  }
  inline long long query(long long rt, long long l, long long r, long long L,
                         long long R, long long dep) {
    if (L <= l && r <= R) return val[rt];
    long long ans = 0;
    if (L <= ((l + r) >> 1)) {
      if (rev[dep])
        ans += query(rt << 1 | 1, l, ((l + r) >> 1), L, R, dep - 1);
      else
        ans += query(rt << 1, l, ((l + r) >> 1), L, R, dep - 1);
    }
    if (R > ((l + r) >> 1)) {
      if (rev[dep])
        ans += query(rt << 1, ((l + r) >> 1) + 1, r, L, R, dep - 1);
      else
        ans += query(rt << 1 | 1, ((l + r) >> 1) + 1, r, L, R, dep - 1);
    }
    return ans;
  }
} t;
signed main() {
  scanf("%lld %lld", &n, &m);
  t.build(1, 1, (1ll << n));
  for (long long i = 1; i <= m; i++) {
    scanf("%lld", &opt);
    if (opt == 1)
      scanf("%lld %lld", &x, &k), t.update(1, 1, (1ll << n), x, k, n);
    if (opt == 2) {
      scanf("%lld", &k);
      for (long long yny = 0; yny <= k; yny++) rev[yny] ^= 1;
    }
    if (opt == 3) scanf("%lld", &k), rev[k + 1] ^= 1;
    if (opt == 4)
      scanf("%lld %lld", &l, &r),
          printf("%lld\n", t.query(1, 1, (1ll << n), l, r, n));
  }
}
