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;
template <typename T>
inline void ckmx(T &x, T y) {
  x = max(x, y);
}
template <typename T>
inline void ckmi(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - 48;
  x *= f;
}
int a[600010];
struct SEG {
  int nown;
  long long val[600010];
  int lc[600010], rc[600010], tag[600010], d[600010];
  void init(int x, int l, int r, int dd) {
    tag[x] = 0;
    d[x] = dd;
    if (l == r)
      val[x] = a[l], lc[x] = 0, rc[x] = 0;
    else {
      int mid = (l + r) / 2;
      ++nown;
      lc[x] = nown;
      init(lc[x], l, mid, dd - 1);
      ++nown;
      rc[x] = nown;
      init(rc[x], mid + 1, r, dd - 1);
      val[x] = val[lc[x]] + val[rc[x]];
    }
  }
  void pd(int x) {
    if (tag[x] & (1 << d[x])) swap(lc[x], rc[x]);
    tag[lc[x]] ^= tag[x];
    tag[rc[x]] ^= tag[x];
    tag[x] = 0;
  }
  void md(int x, int l, int r, int p, int v) {
    if (tag[x]) pd(x);
    if (l == r)
      val[x] = v;
    else {
      int mid = (l + r) / 2;
      if (p <= mid)
        md(lc[x], l, mid, p, v);
      else
        md(rc[x], mid + 1, r, p, v);
      val[x] = val[lc[x]] + val[rc[x]];
    }
  }
  long long qy(int x, int l, int r, int ll, int rr) {
    if (tag[x]) pd(x);
    if (ll <= l && r <= rr)
      return val[x];
    else {
      int mid = (l + r) / 2;
      long long rt = 0;
      if (ll <= mid) rt += qy(lc[x], l, mid, ll, rr);
      if (rr >= mid + 1) rt += qy(rc[x], mid + 1, r, ll, rr);
      return rt;
    }
  }
} seg;
int n, q;
int main() {
  read(n);
  read(q);
  for (int i = 1; i <= (1 << n); ++i) read(a[i]);
  seg.nown = 1;
  seg.init(1, 1, 1 << n, n);
  while (q--) {
    int op;
    read(op);
    if (op == 1) {
      int x, k;
      read(x);
      read(k);
      seg.md(1, 1, 1 << n, x, k);
    } else if (op == 2) {
      int k;
      read(k);
      seg.tag[1] ^= (1 << (k + 1)) - 1;
    } else if (op == 3) {
      int k;
      read(k);
      seg.tag[1] ^= (1 << (k + 1));
    } else {
      int l, r;
      read(l);
      read(r);
      printf("%lld\n", seg.qy(1, 1, 1 << n, l, r));
    }
  }
  return 0;
}
