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;
using ll = long long;
constexpr int INF = 0x3f3f3f3f;
constexpr ll LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr double EPS = 1e-8;
constexpr int MOD = 1000000007;
constexpr int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
constexpr int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1},
              dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U>
inline bool chmax(T &a, U b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
inline bool chmin(T &a, U b) {
  return a > b ? (a = b, true) : false;
}
struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} iosetup;
template <typename Abelian>
struct BIT {
  BIT(int n, const Abelian UNITY = 0) : n(n), UNITY(UNITY), dat(n, UNITY) {}
  void add(int idx, Abelian val) {
    while (idx < n) {
      dat[idx] += val;
      idx |= idx + 1;
    }
  }
  Abelian sum(int idx) const {
    Abelian res = UNITY;
    --idx;
    while (idx >= 0) {
      res += dat[idx];
      idx = (idx & (idx + 1)) - 1;
    }
    return res;
  }
  Abelian sum(int left, int right) const {
    return left < right ? sum(right) - sum(left) : UNITY;
  }
  Abelian operator[](const int idx) const { return sum(idx, idx + 1); }
  int lower_bound(Abelian val) const {
    if (val <= UNITY) return 0;
    int res = 0, exponent = 1;
    while (exponent <= n) exponent <<= 1;
    for (int mask = exponent >> 1; mask > 0; mask >>= 1) {
      if (res + mask - 1 < n && dat[res + mask - 1] < val) {
        val -= dat[res + mask - 1];
        res += mask;
      }
    }
    return res;
  }

 private:
  int n;
  const Abelian UNITY;
  std::vector<Abelian> dat;
};
int main() {
  int n, q;
  cin >> n >> q;
  BIT<ll> a(1 << n);
  for (int i = (0); i < (1 << n); ++i) {
    int ai;
    cin >> ai;
    a.add(i, ai);
  }
  int x = 0;
  while (q--) {
    int query;
    cin >> query;
    if (query == 1) {
      int i, k;
      cin >> i >> k;
      --i;
      a.add(i ^ x, -a[i ^ x]);
      a.add(i ^ x, k);
    } else if (query == 2) {
      int k;
      cin >> k;
      x ^= (1 << k) - 1;
    } else if (query == 3) {
      int k;
      cin >> k;
      x ^= 1 << k;
    } else if (query == 4) {
      int l, r;
      cin >> l >> r;
      --l;
      ll ans = 0;
      while (l < r) {
        int k = 0;
        while (l + (1 << (k + 1)) <= r && !(l >> k & 1)) ++k;
        ans += a.sum(l ^ (x & ~((1 << k) - 1)),
                     (l ^ (x & ~((1 << k) - 1))) + (1 << k));
        l += 1 << k;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
