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;
  const double PI = acos(-1);
  const long long nax = 1e6 + 10, mod = 1e9 + 7, inf = 1e18;
  struct node {
    vector<long long> par, sz;
    node(long long n) : par(n), sz(n) { iota(par.begin(), par.end(), 0); }
    long long find(long long v) {
      return v == par[v] ? v : par[v] = find(par[v]);
    }
    bool same(long long x, long long y) { return find(x) == find(y); }
    bool merge(long long x, long long y) {
      x = find(x);
      y = find(y);
      if (x == y) return 0;
      sz[x] += sz[y];
      par[y] = x;
      return 1;
    }
    long long get(long long x) { return sz[find(x)]; }
  };
  void solve() {
    long long n, m;
    std::cin >> n >> m;
    std::vector<long long> v{0LL, 1LL << n};
    long long op[m];
    long long x[m], y[m];
    for (long long i = 0; i < m; i++) {
      std::string t;
      std::cin >> t;
      if (t == "block") {
        op[i] = 0;
        std::cin >> x[i] >> y[i];
        y[i]++;
        v.push_back(x[i]);
        v.push_back(y[i]);
      } else {
        op[i] = 1;
        std::cin >> x[i] >> y[i];
      }
    }
    std::sort(v.begin(), v.end());
    v.erase(std::unique(v.begin(), v.end()), v.end());
    long long N = v.size() - 1;
    for (long long i = 0; i < N; i++) {
      long long l = v[i];
      long long r = v[i + 1] - 1;
      long long k = 0;
      while ((l >> k) != (r >> k)) {
        k++;
      }
      if (k > 0) {
        v.push_back(r & ~((1LL << (k - 1)) - 1));
      }
    }
    std::sort(v.begin(), v.end());
    v.erase(std::unique(v.begin(), v.end()), v.end());
    N = v.size() - 1;
    node g(N);
    std::vector<long long> adj[N];
    auto add = [&](long long i, long long j, long long k) {
      long long l = std::max(v[i], v[j] - (1LL << k));
      long long r = std::min(v[i + 1], v[j + 1] - (1LL << k)) - 1;
      if ((~l >> k & 1) || (~r >> k & 1) || (l >> k) != (r >> k)) {
        adj[i].push_back(j);
        adj[j].push_back(i);
      }
    };
    for (long long k = 0; k < n; k++) {
      long long d = 1LL << k;
      for (long long i = 0, j = 0; i < N; i++) {
        while (j < N && v[j + 1] <= v[i] + d) {
          j++;
        }
        while (j < N && v[j + 1] <= v[i + 1] + d) {
          add(i, j, k);
          j++;
        }
        if (j < N && v[j] < v[i + 1] + d) {
          add(i, j, k);
        }
      }
    }
    bool vis[N];
    std::fill(vis, vis + N, true);
    for (long long i = 0; i < m; i++) {
      if (op[i] == 0) {
        long long j = std::lower_bound(v.begin(), v.end(), x[i]) - v.begin();
        while (v[j] < y[i]) {
          vis[j++] = false;
        }
      }
    }
    for (long long i = 0; i < N; i++) {
      if (vis[i]) {
        for (auto j : adj[i]) {
          if (vis[j]) {
            g.merge(i, j);
          }
        }
      }
    }
    long long res[m];
    for (long long i = m - 1; i >= 0; i--) {
      if (op[i] == 0) {
        long long j = std::lower_bound(v.begin(), v.end(), x[i]) - v.begin();
        while (v[j] < y[i]) {
          vis[j] = true;
          for (auto k : adj[j]) {
            if (vis[k]) {
              g.merge(j, k);
            }
          }
          j++;
        }
      } else {
        long long a = std::upper_bound(v.begin(), v.end(), x[i]) - v.begin() - 1;
        long long b = std::upper_bound(v.begin(), v.end(), y[i]) - v.begin() - 1;
        res[i] = g.same(a, b);
      }
    }
    for (long long i = 0; i < m; i++) {
      if (op[i] == 1) {
        std::cout << res[i] << "\n";
      }
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(nullptr);
    cout.tie(nullptr);
    long long tt = 1;
    for (long long i = 1; i <= tt; i++) {
      solve();
    }
    return 0;
  }
  