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 MAX = 1 << 19;
int n, dep, q, a[MAX], rev[19];
struct ST {
  struct Node {
    long long sum;
    Node *lc, *rc;
  } node[MAX], *root, null;
  int tail = 0;
  void build() {
    tail = 0;
    root = build(1, n);
  }
  Node *build(int l, int r) {
    Node *u = node + tail++;
    if (l == r) {
      u->sum = a[l];
      u->lc = u->rc = &null;
      return u;
    }
    int m = (l + r) / 2;
    u->lc = build(l, m);
    u->rc = build(m + 1, r);
    u->sum = u->lc->sum + u->rc->sum;
    return u;
  }
  void set(int x, int v) { set(root, 1, n, dep, x, v); }
  void set(Node *u, int cl, int cr, int dep, int x, int v) {
    if (cl == cr) {
      u->sum = v;
      return;
    }
    int m = (cl + cr) / 2;
    if (!rev[dep]) {
      if (x <= m)
        set(u->lc, cl, m, dep - 1, x, v);
      else
        set(u->rc, m + 1, cr, dep - 1, x, v);
    } else {
      if (x <= m)
        set(u->rc, cl, m, dep - 1, x, v);
      else
        set(u->lc, m + 1, cr, dep - 1, x, v);
    }
    u->sum = u->lc->sum + u->rc->sum;
  }
  long long sum(int l, int r) { return sum(root, 1, n, dep, l, r); }
  long long sum(Node *u, int cl, int cr, int dep, int ql, int qr) {
    if (ql <= cl && cr <= qr) return u->sum;
    int m = (cl + cr) / 2;
    long long ans = 0;
    if (!rev[dep]) {
      if (ql <= m) ans += sum(u->lc, cl, m, dep - 1, ql, qr);
      if (qr > m) ans += sum(u->rc, m + 1, cr, dep - 1, ql, qr);
    } else {
      if (ql <= m) ans += sum(u->rc, cl, m, dep - 1, ql, qr);
      if (qr > m) ans += sum(u->lc, m + 1, cr, dep - 1, ql, qr);
    }
    return ans;
  }
} sg;
int main(void) {
  scanf("%d%d", &dep, &q);
  n = 1 << dep;
  for (int i = (1); i <= (n); ++i) scanf("%d", &a[i]);
  sg.build();
  int op, k, x, l, r;
  while (q--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d", &x, &k);
      sg.set(x, k);
    } else if (op == 2) {
      scanf("%d", &k);
      for (int i = (0); i <= (k); ++i) rev[i] ^= 1;
    } else if (op == 3) {
      scanf("%d", &k);
      if (k < dep) rev[k + 1] ^= 1;
    } else {
      scanf("%d%d", &l, &r);
      printf("%lld\n", sg.sum(l, r));
    }
  }
  return 0;
}
