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;
const long long N = 3 * 1e5 + 100;
long long n, q, a[N], vi[4][20], sum[20][N];
inline long long get(long long x, long long id) {
  for (long long i = id; i < n; i++) {
    if (vi[3][i]) {
      long long len = (1 << (i - id)), tmp = (x - 1) / len + 1;
      if (tmp & 1)
        x += len;
      else
        x -= len;
    }
  }
  for (long long i = id + 1; i <= n; i++) {
    if (vi[2][i]) {
      long long len = (1 << (i - id)), tmp = (x - 1) / len + 1;
      long long mid = len * tmp + len * (tmp - 1) + 1;
      x = mid - x;
    }
  }
  return x;
}
inline long long getsum(long long x) {
  long long ans = 0, last = 0;
  for (long long i = n; i >= 0; i--) {
    if (last + (1 << i) <= x) {
      long long tmp = (last / (1 << i)) + 1;
      ans += sum[i][get(tmp, i)];
      last += (1 << i);
    }
  }
  return ans;
}
signed main() {
  scanf("%lld%lld", &n, &q);
  for (long long i = 1; i <= (1 << n); i++)
    scanf("%lld", &a[i]), sum[0][i] = a[i];
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= (1 << (n - i)); j++)
      sum[i][j] = sum[i - 1][2 * j - 1] + sum[i - 1][2 * j];
  }
  while (q--) {
    long long op;
    scanf("%lld", &op);
    if (op == 1) {
      long long x, k;
      scanf("%lld%lld", &x, &k);
      long long delta = k - sum[0][get(x, 0)];
      for (long long i = 0; i <= n; i++)
        sum[i][get((x - 1) / (1 << i) + 1, i)] += delta;
    }
    if (op == 2 || op == 3) {
      long long k;
      scanf("%lld", &k);
      vi[op][k] ^= 1;
    }
    if (op == 4) {
      long long l, r;
      scanf("%lld%lld", &l, &r);
      printf("%lld\n", getsum(r) - getsum(l - 1));
    }
  }
}
