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 maxn = (1 << 18) + 100;
const int maxnode = maxn << 2;
int n, ai[maxn + 5];
int ls[maxnode + 5], rs[maxnode + 5];
int re[maxnode + 5], sp[maxnode + 5];
long long sum[maxnode + 5];
void Build(int x, int l, int r) {
  re[x] = sp[x] = 0;
  ls[x] = x << 1, rs[x] = x << 1 | 1;
  if (l == r) {
    sum[x] = ai[l];
    return;
  }
  int &lson = ls[x], &rson = rs[x], mid = (l + r) >> 1;
  Build(lson, l, mid);
  Build(rson, mid + 1, r);
  sum[x] = sum[lson] + sum[rson];
}
void Replace(int d, int x, int l, int r, int p, int v) {
  if (l == r) {
    sum[x] = v;
    return;
  }
  int &lson = ls[x], &rson = rs[x], mid = (l + r) >> 1;
  if (re[x] >> d & 1) {
    re[x] ^= (1 << d), sp[x] ^= (1 << d), re[lson] ^= (1 << (d + 1)),
        re[rson] ^= (1 << (d + 1));
  }
  if (sp[x] >> d & 1) {
    sp[x] ^= (1 << d), swap(lson, rson);
  }
  re[lson] ^= re[x], re[rson] ^= re[x], re[x] = 0;
  sp[lson] ^= sp[x], sp[rson] ^= sp[x], sp[x] = 0;
  if (p <= mid)
    Replace(d + 1, lson, l, mid, p, v);
  else
    Replace(d + 1, rson, mid + 1, r, p, v);
  sum[x] = sum[lson] + sum[rson];
}
long long Sum(int d, int x, int l, int r, int ll, int rr) {
  if (ll <= l && r <= rr) return sum[x];
  int &lson = ls[x], &rson = rs[x], mid = (l + r) >> 1;
  long long res = 0;
  if (re[x] >> d & 1) {
    re[x] ^= (1 << d), sp[x] ^= (1 << d), re[lson] ^= (1 << (d + 1)),
        re[rson] ^= (1 << (d + 1));
  }
  if (sp[x] >> d & 1) {
    sp[x] ^= (1 << d), swap(lson, rson);
  }
  re[lson] ^= re[x], re[rson] ^= re[x], re[x] = 0;
  sp[lson] ^= sp[x], sp[rson] ^= sp[x], sp[x] = 0;
  if (ll <= mid) res += Sum(d + 1, lson, l, mid, ll, rr);
  if (mid < rr) res += Sum(d + 1, rson, mid + 1, r, ll, rr);
  return res;
}
int main() {
  int _;
  scanf("%d%d", &n, &_);
  int pn = n;
  n = (1 << n);
  for (int i = 1; i <= n; ++i) scanf("%d", ai + i);
  Build(1, 1, n);
  while (_--) {
    int o, x, k, l, r;
    scanf("%d", &o);
    if (o == 1) {
      scanf("%d%d", &x, &k);
      Replace(0, 1, 1, n, x, k);
    } else if (o == 2) {
      scanf("%d", &k);
      k = pn - k;
      re[1] ^= (1 << k);
    } else if (o == 3) {
      scanf("%d", &k);
      k = pn - 1 - k;
      sp[1] ^= (1 << k);
    } else {
      scanf("%d%d", &l, &r);
      long long ans = Sum(0, 1, 1, n, l, r);
      printf("%lld\n", ans);
    }
  }
  return 0;
}
