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("O3", "unroll-loops")
#pragma GCC target("sse4.2")
using namespace std;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
template <class T, class U>
inline bool chmax(T &x, U y) {
  return x < y ? x = y, 1 : 0;
}
template <class T, class U>
inline bool chmin(T &x, U y) {
  return x > y ? x = y, 1 : 0;
}
template <class T>
inline void sort(T &a) {
  sort((a).begin(), (a).end());
}
template <class T>
inline void rsort(T &a) {
  sort((a).rbegin(), (a).rend());
}
template <class T>
inline void reverse(T &a) {
  reverse((a).begin(), (a).end());
}
template <class T, class U>
inline istream &operator>>(istream &str, pair<T, U> &p) {
  return str >> p.first >> p.second;
}
template <class T>
inline istream &operator>>(istream &str, vector<T> &a) {
  for (auto &i : a) str >> i;
  return str;
}
template <class T>
inline T sorted(T a) {
  sort(a);
  return a;
}
void read() {}
void write() {}
void writeln() { cout << '\n'; }
template <class T, class... U>
void read(T &x, U &...u) {
  cin >> x;
  read(u...);
}
template <class T, class... U>
void write(const T &x, const U &...u) {
  cout << x;
  write(u...);
}
template <class T, class... U>
void writeln(const T &x, const U &...u) {
  cout << x;
  writeln(u...);
}
const long long maxn = 20;
long long n, q;
bool swapped[maxn];
long long tree[1ll << maxn];
long long getLeft(long long p, long long level) {
  if (swapped[level])
    return 2 * p + 1;
  else
    return 2 * p;
}
long long getRight(long long p, long long level) {
  if (swapped[level])
    return 2 * p;
  else
    return 2 * p + 1;
}
void upd(long long pos, long long x, long long level, long long p = 1,
         long long l = 1, long long r = 1ll << n) {
  if (l == r && l == pos) {
    tree[p] = x;
    return;
  }
  if (l > pos || r < pos) return;
  upd(pos, x, level - 1, getLeft(p, level), l, (l + r) >> 1);
  upd(pos, x, level - 1, getRight(p, level), ((l + r) >> 1) + 1, r);
  tree[p] = tree[p * 2] + tree[p * 2 + 1];
}
long long sum(long long myL, long long myR, long long level, long long p = 1,
              long long l = 1, long long r = 1ll << n) {
  if (myL > r || myR < l) return 0;
  if (myL <= l && r <= myR) {
    return tree[p];
  }
  return sum(myL, myR, level - 1, getLeft(p, level), l, (l + r) >> 1) +
         sum(myL, myR, level - 1, getRight(p, level), ((l + r) >> 1) + 1, r);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  read(n, q);
  vector<long long> a(1ll << n);
  read(a);
  for (long long i = 0; i < (1ll << n); ++i) {
    upd(i + 1, a[i], n);
  }
  for (long long i = 0; i < q; ++i) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      read(x, k);
      upd(x, k, n);
    } else if (t == 2) {
      long long k;
      read(k);
      for (long long j = 1; j <= k; ++j) swapped[j] ^= 1;
    } else if (t == 3) {
      long long k;
      read(k);
      swapped[k + 1] ^= 1;
    } else if (t == 4) {
      long long l, r;
      read(l, r);
      writeln(sum(l, r, n));
    }
  }
  return 0;
}
