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 rv[19];
bool sw[19];
struct node {
  int64_t s, e, v, level;
};
node tree[(1 << 19) * 2];
void tree_init(int64_t at, int64_t s, int64_t e, int64_t level) {
  if (0)
    for (int64_t i = level; i < 19; i++) printf(" ");
  if (0) printf("#%ld: [%ld, %ld] (%ld)\n", at, s, e, level);
  tree[at].s = s;
  tree[at].e = e;
  tree[at].v = 0;
  tree[at].level = level;
  if (s != e) {
    int64_t m = (s + e) / 2;
    tree_init(at * 2 + 1, s, m, level - 1);
    tree_init(at * 2 + 2, m + 1, e, level - 1);
  }
}
int64_t re_index(int64_t index) {
  for (int64_t i = 19 - 1; i >= 0; i--) {
    if (rv[i]) {
      int64_t step = (1 << i);
      int64_t at = index / step;
      int64_t nx = at + 1;
      index = at * step + (nx * step - 1 - index);
    }
    if (sw[i]) {
      int64_t step = (1 << i);
      int64_t at = index / step;
      if (at % 2 == 0) {
        index += step;
      } else {
        index -= step;
      }
    }
  }
  return index;
}
void tree_set(int64_t index, int64_t v) {
  index = re_index(index);
  if (0) printf("set a[%ld] = %ld\n", index, v);
  int64_t at = (1 << 19) - 1 + index;
  tree[at].v = v;
  while (at != 0) {
    int64_t parent = (at - 1) / 2;
    tree[parent].v = tree[parent * 2 + 1].v + tree[parent * 2 + 2].v;
    at = parent;
  }
}
int64_t tree_sum(int64_t at, int64_t start, int64_t end, bool reversed) {
  int64_t bounds = (1 << tree[at].level) - 1;
  if (start > end || end < 0 || start > bounds) return 0;
  start = max(start, (int64_t)0L);
  end = min(end, bounds);
  reversed = reversed ^ rv[tree[at].level];
  bool swapped = tree[at].level != 0 && sw[tree[at].level - 1];
  if (0) {
    for (int64_t i = tree[at].level; i < 19; i++) printf(" ");
    printf("at %ld [%ld, %ld], %d %d = %ld\n", at, start, end, reversed,
           swapped, tree[at].v);
  }
  if (start == 0 && end == bounds) return tree[at].v;
  if (tree[at].level == 0) return tree[at].v;
  int64_t hw = (1 << tree[at].level) / 2;
  if (reversed) {
    int64_t a = start;
    int64_t b = end;
    start = bounds - b;
    end = bounds - a;
    return tree_sum(at, start, end, rv[tree[at].level]);
  } else if (swapped) {
    return (tree_sum(at * 2 + 1, start - hw, end - hw, reversed) +
            tree_sum(at * 2 + 2, start, end, reversed));
  } else {
    return (tree_sum(at * 2 + 1, start, end, reversed) +
            tree_sum(at * 2 + 2, start - hw, end - hw, reversed));
  }
}
int64_t n, q, a, t, x, y;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> q;
  tree_init(0, 0, (1 << 19) - 1, 19);
  for (int64_t i = 0; i < (1 << n); i++) {
    cin >> a;
    tree_set(i, a);
  }
  while (q--) {
    cin >> t;
    if (t == 1) {
      cin >> x >> y;
      x -= 1;
      tree_set(x, y);
    }
    if (t == 2) {
      cin >> x;
      rv[x] = !rv[x];
    }
    if (t == 3) {
      cin >> x;
      sw[x] = !sw[x];
    }
    if (t == 4) {
      cin >> x >> y;
      x -= 1;
      y -= 1;
      cout << tree_sum(0, x, y, false) << endl;
    }
    if (0) {
      for (int64_t i = 0; i < (1 << n); i++) {
        cout << tree[(1 << 19) - 1 + re_index(i)].v << ' ';
      }
      cout << endl;
    }
  }
}
