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 int LOGN = 20, N = 1 << LOGN;
bool flipped[LOGN];
struct node {
  int dt, tp;
  long long s;
  node *ch[2];
  node(int _dt, int _tp) {
    ch[0] = ch[1] = NULL;
    dt = _dt;
    tp = _tp;
    s = 0;
  }
  void merge() { s = ch[0]->s + ch[1]->s; }
} * root;
long long query(int L, int R, int depth, node *t) {
  if (L > t->tp || R < 1) return 0;
  if (L <= 1 && R >= t->tp) {
    return t->s;
  }
  bool f = flipped[depth - 1] ^ 1;
  int s = t->ch[f ^ 1]->tp;
  long long left = query(L, R, depth - 1, t->ch[f ^ 1]);
  long long right = query(max(1, L - s), R - s, depth - 1, t->ch[f]);
  return left + right;
}
void update(int sz, int x, int value, int depth, node *&t) {
  if (!t) t = new node(depth, sz);
  if (x > t->tp || x < 1) return;
  if (t->tp == 1) {
    t->s = value;
  } else {
    bool f = flipped[depth - 1] ^ 1;
    update((sz + 1) / 2, x, value, depth - 1, t->ch[f ^ 1]);
    update(sz / 2, x - t->ch[f ^ 1]->tp, value, depth - 1, t->ch[f]);
    t->merge();
  }
}
int main() {
  int logn, q;
  cin >> logn >> q;
  int n = 1 << logn;
  for (int i = 1, a; i <= n; ++i) {
    cin >> a;
    update(n, i, a, logn, root);
  }
  while (q--) {
    int TYPE, x, k, l, r;
    cin >> TYPE;
    if (TYPE == 1) {
      cin >> x >> k;
      update(n, x, k, logn, root);
    }
    if (TYPE == 2) {
      cin >> k;
      for (int i = 0; i < k; ++i) flipped[i] ^= 1;
    }
    if (TYPE == 3) {
      cin >> k;
      flipped[k] ^= 1;
    }
    if (TYPE == 4) {
      cin >> l >> r;
      cout << query(l, r, logn, root) << '\n';
    }
  }
}
