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 = 3e5 + 5;
const long long mo = 1e9 + 7;
int n, m, k, nn;
template <typename T>
inline void read(T &X) {
  X = 0;
  int w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  if (w) X = -X;
}
long long sum[maxn << 2], fg[maxn << 2];
int rvs[maxn << 2], ls[maxn << 2], rs[maxn << 2];
long long a[maxn];
void pu(int rt) { sum[rt] = sum[ls[rt]] + sum[rs[rt]]; }
void build(int rt, int l, int r) {
  fg[rt] = rvs[rt] = 0;
  if (l == r) {
    sum[rt] = a[l];
    return;
  }
  ls[rt] = (rt << 1);
  rs[rt] = (rt << 1 | 1);
  int mid = (l + r) >> 1;
  build(ls[rt], l, mid);
  build(rs[rt], mid + 1, r);
  pu(rt);
}
void pd(int rt, int x) {
  if (rvs[rt] & x) {
    if (fg[rt] & x) {
      fg[rt] ^= x;
    } else {
      swap(ls[rt], rs[rt]);
    }
    rvs[rt] ^= x;
    rvs[ls[rt]] ^= (x >> 1);
    rvs[rs[rt]] ^= (x >> 1);
  } else {
    if (fg[rt] & x) {
      swap(ls[rt], rs[rt]);
      fg[rt] ^= x;
    }
  }
  rvs[ls[rt]] ^= rvs[rt];
  rvs[rs[rt]] ^= rvs[rt];
  fg[ls[rt]] ^= fg[rt];
  fg[rs[rt]] ^= fg[rt];
  rvs[rt] = fg[rt] = 0;
}
void update(int rt, int l, int r, int pos, int v) {
  if (l == r) {
    sum[rt] = v;
    return;
  }
  pd(rt, r - l + 1);
  int mid = (l + r) >> 1;
  if (pos <= mid) update(ls[rt], l, mid, pos, v);
  if (pos > mid) update(rs[rt], mid + 1, r, pos, v);
  pu(rt);
}
long long query(int rt, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) {
    return sum[rt];
  }
  pd(rt, r - l + 1);
  int mid = (l + r) >> 1;
  long long ans = 0;
  if (ql <= mid) ans += query(ls[rt], l, mid, ql, qr);
  if (qr > mid) ans += query(rs[rt], mid + 1, r, ql, qr);
  return ans;
}
void solve() {
  read(n);
  read(m);
  n = (1 << n);
  for (int i = (1); i <= (n); i++) {
    read(a[i]);
  }
  build(1, 1, n);
  while (m--) {
    int id, x, y;
    read(id);
    if (id == 1) {
      read(x);
      read(y);
      update(1, 1, n, x, y);
    } else if (id == 2) {
      read(x);
      rvs[1] ^= (1 << x);
    } else if (id == 3) {
      read(k);
      k++;
      fg[1] ^= (1 << k);
    } else {
      read(x);
      read(y);
      long long ans = query(1, 1, n, x, y);
      printf("%lld\n", ans);
    }
  }
}
int main() {
  int T = 1, cas = 1;
  while (T--) {
    solve();
  }
  return 0;
}
