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>
#pragma GCC optimize("O2")
using namespace std;
const int N = 1e6 + 10;
const long long mod = 1e9 + 7;
const long long mod2 = 998244353;
const long long inf = 8e18;
const int LOG = 22;
long long pw(long long a, long long b, long long M) {
  return (!b ? 1
             : (b & 1 ? (a * pw(a * a % M, b / 2, M)) % M
                      : pw(a * a % M, b / 2, M)));
}
long long L[4 * N], R[4 * N], seg[4 * N], second[4 * LOG], H[4 * N], A[4 * N];
void B(int v, int tl, int tr) {
  if (tl == tr) {
    seg[v] = A[tl];
    return;
  }
  int mid = (tl + tr) >> 1;
  L[v] = 2 * v;
  R[v] = 2 * v + 1;
  H[R[v]] = H[v] + 1;
  H[L[v]] = H[v] + 1;
  B(L[v], tl, mid);
  B(R[v], mid + 1, tr);
  seg[v] = (seg[L[v]] + seg[R[v]]);
}
void U(int v, int tl, int tr, int x, int now) {
  if (second[H[v]] % 2) {
    swap(L[v], R[v]);
  }
  if (x < tl || x > tr) {
    if (second[H[v]] % 2) {
      swap(L[v], R[v]);
    }
    return;
  }
  if (tl == x && tr == x) {
    seg[v] = now;
    if (second[H[v]] % 2) {
      swap(L[v], R[v]);
    }
    return;
  }
  int mid = (tl + tr) >> 1;
  U(L[v], tl, mid, x, now);
  U(R[v], mid + 1, tr, x, now);
  seg[v] = (seg[L[v]] + seg[R[v]]);
  if (second[H[v]] % 2) {
    swap(L[v], R[v]);
  }
}
long long G(int v, int tl, int tr, int l, int r) {
  if (second[H[v]] % 2) {
    swap(L[v], R[v]);
  }
  if (tl == l && tr == r) {
    if (second[H[v]] % 2) {
      swap(L[v], R[v]);
    }
    return seg[v];
  }
  int mid = (tl + tr) >> 1;
  long long ans = 0;
  if (l <= mid) {
    ans = G(L[v], tl, mid, l, min(r, mid));
  }
  if (r > mid) {
    ans += G(R[v], mid + 1, tr, max(mid + 1, l), r);
  }
  if (second[H[v]] % 2) {
    swap(L[v], R[v]);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, q;
  cin >> n >> q;
  int n2 = (1 << n);
  for (int i = 1; i <= n2; i++) {
    cin >> A[i];
  }
  B(1, 1, n2);
  while (q--) {
    int w;
    cin >> w;
    if (w == 1) {
      long long x, now;
      cin >> x >> now;
      U(1, 1, n2, x, now);
    } else if (w == 2) {
      int k;
      cin >> k;
      for (int i = n - k; i <= n + 1; i++) {
        second[i]++;
      }
    } else if (w == 3) {
      int k;
      cin >> k;
      second[n - k - 1]++;
    } else {
      int l, r;
      cin >> l >> r;
      cout << G(1, 1, n2, l, r) << "\n";
    }
  }
  return 0;
}
