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", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
const long long MOD2 = (long long)1000000007 * (long long)1000000007;
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 1000000007) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
int N, a[(270135)];
long long T[(270135) * 4];
bool isR[20];
void Up(int rt, int l, int r) { T[rt] = T[(rt << 1)] + T[(rt << 1 | 1)]; }
void build(int rt, int l, int r) {
  if (l == r) {
    T[rt] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build((rt << 1), l, mid);
  build((rt << 1 | 1), mid + 1, r);
  Up(rt, l, r);
}
void upt(int rt, int l, int r, int dep, int p, int val) {
  if (l == r) {
    T[rt] = val;
    return;
  }
  int mid = (l + r) / 2;
  if (p <= mid) {
    upt(isR[dep] ? (rt << 1 | 1) : (rt << 1), l, mid, dep - 1, p, val);
  } else {
    upt(isR[dep] ? (rt << 1) : (rt << 1 | 1), mid + 1, r, dep - 1, p, val);
  }
  Up(rt, l, r);
}
long long qy(int rt, int l, int r, int L, int R, int dep) {
  if (R < l || r < L) return 0;
  if (L <= l && r <= R) {
    return T[rt];
  }
  int mid = (l + r) / 2;
  return qy(isR[dep] ? (rt << 1 | 1) : (rt << 1), l, mid, L, R, dep - 1) +
         qy(isR[dep] ? (rt << 1) : (rt << 1 | 1), mid + 1, r, L, R, dep - 1);
}
void fmain(int tid) {
  scanf("%d%d", &n, &m);
  N = 1 << n;
  for (int(i) = 1; (i) <= (int)(N); (i)++) scanf("%d", a + i);
  build(1, 1, N);
  for (int(i) = 1; (i) <= (int)(m); (i)++) {
    int tp, x;
    scanf("%d", &tp);
    if (tp == 1) {
      scanf("%d%d", &x, &k);
      upt(1, 1, N, n, x, k);
    } else if (tp == 2) {
      scanf("%d", &k);
      for (int j = 0; j <= k; j++) isR[j] ^= 1;
    } else if (tp == 3) {
      scanf("%d", &k);
      isR[k + 1] ^= 1;
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%lld\n", qy(1, 1, N, l, r, n));
    }
  }
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}
