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;
constexpr long long mod = 1000000007;
const long long INF = mod * mod;
const long double eps = 1e-8;
const long double pi = acosl(-1.0);
long long mod_pow(long long x, long long n, long long m = mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % m;
    x = x * x % m;
    n >>= 1;
  }
  return res;
}
struct modint {
  long long n;
  modint() : n(0) { ; }
  modint(long long m) : n(m) {
    if (n >= mod)
      n %= mod;
    else if (n < 0)
      n = (n % mod + mod) % mod;
  }
  operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint& a, modint b) {
  a.n += b.n;
  if (a.n >= mod) a.n -= mod;
  return a;
}
modint operator-=(modint& a, modint b) {
  a.n -= b.n;
  if (a.n < 0) a.n += mod;
  return a;
}
modint operator*=(modint& a, modint b) {
  a.n = ((long long)a.n * b.n) % mod;
  return a;
}
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, long long n) {
  if (n == 0) return modint(1);
  modint res = (a * a) ^ (n / 2);
  if (n % 2) res = res * a;
  return res;
}
long long inv(long long a, long long p) {
  return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }
const int max_n = 1 << 19;
modint fact[max_n], factinv[max_n];
void init_f() {
  fact[0] = modint(1);
  for (int i = 0; i < max_n - 1; i++) {
    fact[i + 1] = fact[i] * modint(i + 1);
  }
  factinv[max_n - 1] = modint(1) / fact[max_n - 1];
  for (int i = max_n - 2; i >= 0; i--) {
    factinv[i] = factinv[i + 1] * modint(i + 1);
  }
}
modint comb(int a, int b) {
  if (a < 0 || b < 0 || a < b) return 0;
  return fact[a] * factinv[b] * factinv[a - b];
}
int gcd(int a, int b) {
  if (a < b) swap(a, b);
  while (b) {
    int r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long node[1 << 19];
void solve() {
  int n;
  cin >> n;
  int q;
  cin >> q;
  int m = (1 << n);
  vector<int> a(m);
  for (int i = 0; i < m; i++) {
    cin >> a[i];
    node[i + m - 1] = a[i];
  }
  for (int i = m - 1 - 1; i >= 0; i--) {
    node[i] = node[2 * i + 1] + node[2 * i + 2];
  }
  int ss = 0;
  auto calc = [&](int ri) -> long long {
    if (ri == m) return node[0];
    long long res = 0;
    int loc = 0;
    for (int i = n - 1; i >= 0; i--) {
      bool b = ss & (1 << i);
      if (ri & (1 << i)) {
        res += node[2 * loc + 1 + b];
        loc = 2 * loc + 2 - b;
      } else {
        loc = 2 * loc + 1 + b;
      }
    }
    return res;
  };
  for (int i = 0; i < q; i++) {
    int typ;
    cin >> typ;
    if (typ == 1) {
      int x, k;
      cin >> x >> k;
      x--;
      x ^= ss;
      x += m - 1;
      node[x] = k;
      while (x > 0) {
        x = (x - 1) / 2;
        node[x] = node[2 * x + 1] + node[2 * x + 2];
      }
    } else if (typ == 2) {
      int x;
      cin >> x;
      ss ^= (1 << x) - 1;
    } else if (typ == 3) {
      int x;
      cin >> x;
      ss ^= (1 << x);
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      long long ans = calc(r + 1) - calc(l);
      cout << ans << "\n";
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
