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("Ofast")
using namespace std;
long long a[300000];
long long sum[600000], val[600000], lc[600000], rc[600000];
long long n, q;
void pd(long long x, long long l, long long r, long long d) {
  if (l != r) {
    if (val[x] & (1 << d)) {
      swap(lc[x], rc[x]);
    }
    val[lc[x]] ^= val[x];
    val[rc[x]] ^= val[x];
    val[x] = 0;
  }
}
void build(long long x, long long l, long long r) {
  if (l == r) {
    sum[x] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  lc[x] = x << 1;
  rc[x] = (x << 1) | 1;
  build(x << 1, l, mid);
  build((x << 1) | 1, mid + 1, r);
  sum[x] = sum[lc[x]] + sum[rc[x]];
}
void ch(long long x, long long l, long long r, long long t, long long v,
        long long d) {
  pd(x, l, r, d);
  if (l > t || r < t) return;
  if (l == r) {
    sum[x] = v;
    return;
  }
  long long mid = (l + r) >> 1;
  ch(lc[x], l, mid, t, v, d - 1);
  ch(rc[x], mid + 1, r, t, v, d - 1);
  sum[x] = sum[lc[x]] + sum[rc[x]];
}
long long get(long long x, long long l, long long r, long long tl, long long tr,
              long long d) {
  pd(x, l, r, d);
  if (l > tr || tl > r) return 0;
  if (l >= tl && r <= tr) return sum[x];
  long long mid = (l + r) >> 1;
  return get(lc[x], l, mid, tl, tr, d - 1) +
         get(rc[x], mid + 1, r, tl, tr, d - 1);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  long long m = n;
  n = 1 << n;
  long long x, y;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  for (long long i = 1; i <= q; i++) {
    cin >> x;
    if (x == 1) {
      cin >> x >> y;
      ch(1, 1, n, x, y, m);
    } else if (x == 2) {
      cin >> x;
      x++;
      val[1] ^= (1 << x) - 1;
    } else if (x == 3) {
      cin >> x;
      x++;
      val[1] ^= (1 << x);
    } else {
      cin >> x >> y;
      cout << get(1, 1, n, x, y, m) << '\n';
    }
  }
  return 0;
}
