description: <image>
  
  As mentioned previously William really likes playing video games. In one of his favorite games, the player character is in a universe where every planet is designated by a binary number from 0 to 2^n - 1. On each planet, there are gates that allow the player to move from planet i to planet j if the binary representations of i and j differ in exactly one bit.
  
  William wants to test you and see how you can handle processing the following queries in this game universe:
  
    * Destroy planets with numbers from l to r inclusively. These planets cannot be moved to anymore.
    * Figure out if it is possible to reach planet b from planet a using some number of planetary gates. It is guaranteed that the planets a and b are not destroyed. 
  
  Input
  
  The first line contains two integers n, m (1 ≤ n ≤ 50, 1 ≤ m ≤ 5 ⋅ 10^4), which are the number of bits in binary representation of each planets' designation and the number of queries, respectively.
  
  Each of the next m lines contains a query of two types:
  
  block l r — query for destruction of planets with numbers from l to r inclusively (0 ≤ l ≤ r < 2^n). It's guaranteed that no planet will be destroyed twice.
  
  ask a b — query for reachability between planets a and b (0 ≤ a, b < 2^n). It's guaranteed that planets a and b hasn't been destroyed yet.
  
  Output
  
  For each query of type ask you must output "1" in a new line, if it is possible to reach planet b from planet a and "0" otherwise (without quotation marks).
  
  Examples
  
  Input
  
  
  3 3
  ask 0 7
  block 3 6
  ask 0 7
  
  
  Output
  
  
  1
  0
  
  
  Input
  
  
  6 10
  block 12 26
  ask 44 63
  block 32 46
  ask 1 54
  block 27 30
  ask 10 31
  ask 11 31
  ask 49 31
  block 31 31
  ask 2 51
  
  
  Output
  
  
  1
  1
  0
  0
  1
  0
  
  Note
  
  The first example test can be visualized in the following way:
  
  <image>
  
  Response to a query ask 0 7 is positive.
  
  Next after query block 3 6 the graph will look the following way (destroyed vertices are highlighted):
  
  <image>
  
  Response to a query ask 0 7 is negative, since any path from vertex 0 to vertex 7 must go through one of the destroyed vertices.
solution: #include <bits/stdc++.h>
  using namespace std;
  template <typename T>
  inline void ckmax(T &x, T y) {
    x = x < y ? y : x;
    return;
  }
  template <typename T>
  inline void ckmin(T &x, T y) {
    x = x < y ? x : y;
    return;
  }
  inline long long read() {
    long long res = 0, f = 1;
    char k;
    while (!isdigit(k = getchar()))
      if (k == '-') f = -1;
    while (isdigit(k)) res = res * 10 + k - '0', k = getchar();
    return res * f;
  }
  namespace Output {
  char OPUT[100];
  inline void print(int x) {
    if (!x) return putchar('0'), putchar('\n'), void();
    if (x < 0) putchar('-'), x = -x;
    int cnt = 0;
    while (x) OPUT[++cnt] = x % 10, x /= 10;
    for (register int i = cnt; i >= 1; --i) putchar(OPUT[i] + '0');
    putchar('\n');
    return;
  }
  }  // namespace Output
  using namespace Output;
  const int M = 5e4 + 10, N = M << 6;
  int ls[N], rs[N], fa[N], cnt, n, m, ans[M], qcnt, rt, opt[M], tim[N];
  char s[10];
  long long y[M], x[M];
  inline int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
  inline void push_down(int p) {
    if (!ls[p]) ls[p] = ++cnt, fa[cnt] = cnt;
    if (!rs[p]) rs[p] = ++cnt, fa[cnt] = cnt;
    if (tim[p]) tim[rs[p]] = tim[ls[p]] = tim[p], tim[p] = 0;
    return;
  }
  vector<pair<int, int> > upd[M];
  inline void merge(int x, int y) {
    if (ls[x] + rs[x] + ls[y] + rs[y] == 0)
      return upd[min(tim[x], tim[y])].push_back(make_pair(x, y)), void();
    else if (ls[x] + rs[x] == 0)
      push_down(y), merge(x, ls[y]), merge(x, rs[y]);
    else if (ls[y] + rs[y] == 0)
      push_down(x), merge(ls[x], y), merge(rs[x], y);
    else
      push_down(x), push_down(y), merge(ls[x], ls[y]), merge(rs[x], rs[y]);
    return;
  }
  inline void push(int p, long long l, long long r, long long st, long long ed,
                   int tmp) {
    if (st <= l && r <= ed) return tim[p] = tmp, void();
    long long mid = (l + r) >> 1;
    push_down(p);
    if (st <= mid) push(ls[p], l, mid, st, ed, tmp);
    if (ed > mid) push(rs[p], mid + 1, r, st, ed, tmp);
    return;
  }
  inline int getid(int p, long long l, long long r, long long pos) {
    if (ls[p] + rs[p] == 0) return p;
    long long mid = (l + r) >> 1;
    if (pos <= mid)
      return getid(ls[p], l, mid, pos);
    else
      return getid(rs[p], mid + 1, r, pos);
  }
  signed main() {
    n = read();
    m = read();
    tim[fa[rt = cnt = 1] = 1] = m + 1;
    long long U = 1ll << n;
    --U;
    for (register int i = 1; i <= m; ++i) {
      scanf("%s", s + 1);
      x[i] = read(), y[i] = read();
      if (s[1] == 'b') push(rt, 0, U, x[i], y[i], i), opt[i] = 1;
    }
    for (register int i = 1; i <= cnt; ++i)
      if (ls[i] + rs[i]) merge(ls[i], rs[i]);
    for (auto t : upd[m + 1]) fa[find(t.first)] = find(t.second);
    for (register int i = m; i >= 1; --i) {
      if (opt[i])
        for (auto t : upd[i]) fa[find(t.first)] = find(t.second);
      else
        ans[++qcnt] = find(getid(rt, 0, U, x[i])) == find(getid(rt, 0, U, y[i]));
    }
    reverse(ans + 1, ans + qcnt + 1);
    for (register int i = 1; i <= qcnt; ++i) print(ans[i]);
    return 0;
  }
  