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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f == 1 ? x : -x;
}
inline void print(long long x) {
  if (x < 0) x = ~x + 1, putchar('-');
  if (x >= 10) print(x / 10);
  putchar((x % 10) | 48);
}
long long n, q;
long long a[1000010];
long long seg[4000010];
long long rev[200];
void pushup(long long p) { seg[p] = seg[p << 1] + seg[p << 1 | 1]; }
long long ls(long long p, long long k) {
  if (rev[k])
    return p << 1 | 1;
  else
    return p << 1;
}
long long rs(long long p, long long k) {
  if (rev[k])
    return p << 1;
  else
    return p << 1 | 1;
}
void build(long long p, long long l, long long r, long long po) {
  if (l == r) {
    seg[p] = a[l];
    return;
  }
  long long mid = l + r >> 1;
  build(ls(p, po), l, mid, po - 1);
  build(rs(p, po), mid + 1, r, po - 1);
  pushup(p);
}
void update(long long p, long long l, long long r, long long po, long long pos,
            long long k) {
  if (l == r) {
    seg[p] = k;
    return;
  }
  long long mid = l + r >> 1;
  if (pos <= mid)
    update(ls(p, po), l, mid, po - 1, pos, k);
  else
    update(rs(p, po), mid + 1, r, po - 1, pos, k);
  pushup(p);
}
long long query(long long p, long long l, long long r, long long L, long long R,
                long long po) {
  if (l >= L && r <= R) return seg[p];
  long long mid = l + r >> 1;
  long long res = 0;
  if (L <= mid) res += query(ls(p, po), l, mid, L, R, po - 1);
  if (R > mid) res += query(rs(p, po), mid + 1, r, L, R, po - 1);
  return res;
}
signed main() {
  n = read(), q = read();
  for (long long i = 1; i <= (1 << n); ++i) a[i] = read();
  build(1, 1, 1 << n, n - 1);
  while (q--) {
    long long op = read();
    switch (op) {
      case 1: {
        long long x = read(), k = read();
        update(1, 1, (1 << n), n - 1, x, k);
        break;
      }
      case 2: {
        long long k = read();
        for (long long j = k - 1; j >= 0; --j) rev[j] ^= 1;
        break;
      }
      case 3: {
        long long k = read();
        rev[k] ^= 1;
        break;
      }
      case 4: {
        long long l = read(), r = read();
        printf("%lld\n", query(1, 1, 1 << n, l, r, n - 1));
        break;
      }
    }
  }
  return 0;
}
