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;
bool sw[20], rev[20];
int n, q;
const int MAXN = (1 << 19) + 10;
long long A[MAXN];
long long sum[MAXN];
long long B[MAXN];
bool d[MAXN];
void updbit(int pos, int val) {
  while (pos <= (1 << n)) {
    B[pos] += val;
    pos += pos & -pos;
  }
}
long long querybit(int pos) {
  long long s = 0;
  while (pos) {
    s += B[pos];
    pos -= pos & -pos;
  }
  return s;
}
void init(int node, int li, int ls) {
  if (li == ls) {
    sum[node] = A[li];
  } else {
    int mid = (li + ls) / 2;
    init(2 * node, li, mid);
    init(2 * node + 1, mid + 1, ls);
    sum[node] = sum[2 * node] + sum[2 * node + 1];
  }
}
void upd(int node, int li, int ls, int i, int j, int k, int h = 0) {
  if (sw[h]) {
    updbit((1 << h), 1);
    updbit((1 << (h + 1)), -1);
    sw[h] = false;
  }
  if (rev[h]) {
    updbit((1 << h), 1);
    updbit((1 << (h + 1)), -1);
    rev[h] = false;
    if (li != ls) {
      rev[h + 1] ^= 1;
    }
  }
  if (querybit(node) % 2) {
    d[node] ^= 1;
    updbit(node, 1);
    updbit(node + 1, -1);
  }
  if (i > ls || j < li) return;
  if (li >= i && ls <= j) {
    sum[node] = k;
    return;
  }
  int mid = (li + ls) / 2;
  upd(2 * node + d[node], li, mid, i, j, k, h + 1);
  upd(2 * node + !d[node], mid + 1, ls, i, j, k, h + 1);
  sum[node] = sum[2 * node] + sum[2 * node + 1];
}
long long query(int node, int li, int ls, int i, int j, int h = 0) {
  if (sw[h]) {
    updbit((1 << h), 1);
    updbit((1 << (h + 1)), -1);
    sw[h] = false;
  }
  if (rev[h]) {
    updbit((1 << h), 1);
    updbit((1 << (h + 1)), -1);
    rev[h] = false;
    if (li != ls) {
      rev[h + 1] ^= 1;
    }
  }
  if (querybit(node) % 2) {
    d[node] ^= 1;
    updbit(node, 1);
    updbit(node + 1, -1);
  }
  if (i > ls || j < li) return 0;
  if (li >= i && ls <= j) return sum[node];
  int mid = (li + ls) / 2;
  long long left = query(2 * node + d[node], li, mid, i, j, h + 1);
  long long right = query(2 * node + !d[node], mid + 1, ls, i, j, h + 1);
  return left + right;
}
void replaceOp() {
  int x, k;
  scanf("%d %d", &x, &k);
  upd(1, 1, (1 << n), x, x, k);
}
void reverseOp() {
  int k;
  scanf("%d", &k);
  k = n - k;
  rev[k] ^= 1;
}
void swapOp() {
  int k;
  scanf("%d", &k);
  k = n - k - 1;
  sw[k] ^= 1;
}
long long sumOp() {
  int l, r;
  scanf("%d %d", &l, &r);
  return query(1, 1, (1 << n), l, r);
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= (1 << n); i++) scanf("%lld", &A[i]);
  init(1, 1, (1 << n));
  while (q--) {
    int qi;
    scanf("%d", &qi);
    if (qi == 1) {
      replaceOp();
    } else if (qi == 2) {
      reverseOp();
    } else if (qi == 3) {
      swapOp();
    } else {
      printf("%lld\n", sumOp());
    }
  }
  return 0;
}
