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 int INF = 0x3f3f3f3f;
  const int MOD = (1 << 23) - 1;
  int n, m, cntseg;
  int fa[20000000];
  map<pair<long long, long long>, int> segid;
  map<long long, int> lg;
  int root(int x) { return fa[x] = (fa[x] == x ? x : root(fa[x])); }
  void merge(int u, int v) { fa[root(u)] = root(v); }
  set<pair<long long, long long> > alive;
  set<pair<pair<long long, long long>, int> > alive2;
  vector<pair<long long, long long> > segs;
  void get(long long A, long long B, long long L = 0, long long R = 1ll << n) {
    if (R <= B && L >= A) {
      segs.push_back(make_pair(L, R - 1));
      return;
    }
    if (R <= A || L >= B) return;
    long long mid = (L + R) >> 1;
    get(A, B, L, mid);
    get(A, B, mid, R);
  }
  void add(long long l, long long r) {
    alive.insert(make_pair(l, r));
    segid[make_pair(l, r)] = ++cntseg;
    auto ite2 = alive2.lower_bound(make_pair(make_pair(r + 1, 0), 0));
    if (ite2 != alive2.begin() && prev(ite2)->first.first <= l &&
        prev(ite2)->first.second >= r)
      merge(cntseg, prev(ite2)->second);
    else
      alive2.insert(make_pair(make_pair(l, r), cntseg));
    int now = cntseg;
    int d_ = lg[r - l + 1];
    for (int d = d_; d <= n - 1; ++d) {
      l ^= 1ll << d;
      r ^= 1ll << d;
      auto ite = alive.lower_bound(make_pair(r + 1, 0));
      bool z = 0;
      if (ite != alive.begin()) {
        --ite;
        if (ite->first <= l && ite->second >= r) {
          merge(now, segid[*ite]);
          z = 1;
        }
        ite++;
      }
      if (!z && ite != alive.begin() && prev(ite)->second >= l) {
        auto ite3 = alive2.lower_bound(make_pair(make_pair(r + 1, 0), 0));
        if (ite3 != alive2.begin() && prev(ite3)->first.first <= l &&
            prev(ite3)->first.second >= r)
          merge(cntseg, prev(ite3)->second);
        else {
          auto ite = alive2.lower_bound(make_pair(make_pair(l, 0), 0));
          while (ite != alive2.end()) {
            if (ite->first.first > r) break;
            merge(cntseg, ite->second);
            ite = alive2.erase(ite);
          }
          alive2.insert(make_pair(make_pair(l, r), cntseg));
        }
      }
      l ^= 1ll << d;
      r ^= 1ll << d;
    }
  }
  int main() {
    scanf("%d%d", &n, &m);
    vector<pair<string, pair<long long, long long> > > op;
    vector<int> ans;
    map<long long, int> lp;
    for (int i = 0; i < 60; ++i) lg[1ll << i] = i;
    lp[0] = 1;
    int cnt_ask = 0;
    for (int i = 1; i <= m; ++i) {
      string s;
      long long l, r;
      cin >> s;
      scanf("%lld%lld", &l, &r);
      op.push_back(make_pair(s, make_pair(l, r)));
      cnt_ask += (s[0] == 'a');
      if (s == "block") lp[l] = lp[r + 1] = 1;
    }
    if (cnt_ask == 0) {
      return 0;
    }
    if (cnt_ask == 1 && op[0].first == "ask") {
      cout << 1 << endl;
      return 0;
    }
    lp[1ll << n] = 1;
    for (auto ite = lp.begin(); next(ite) != lp.end(); ite++) {
      segs.clear();
      get(ite->first, next(ite)->first);
      for (auto it : segs) alive.insert(it);
    }
    for (int i = 0; i < 20000000; ++i) fa[i] = i;
    reverse(op.begin(), op.end());
    for (auto it : op) {
      if (it.first == "block") {
        segs.clear();
        get(it.second.first, it.second.second + 1);
        for (auto it : segs) alive.erase(it);
      }
    }
    vector<pair<long long, long long> > Tmp;
    for (auto ite = alive.begin(); ite != alive.end(); ite++) Tmp.push_back(*ite);
    alive.clear();
    for (auto it : Tmp) {
      add(it.first, it.second);
    }
    for (auto it : op) {
      if (it.first == "ask") {
        long long a, b;
        a = it.second.first;
        b = it.second.second;
        auto itea = alive.lower_bound(make_pair(a + 1, 0));
        itea--;
        auto iteb = alive.lower_bound(make_pair(b + 1, 0));
        iteb--;
        ans.push_back(root(segid[*itea]) == root(segid[*iteb]));
      }
      if (it.first == "block") {
        segs.clear();
        get(it.second.first, it.second.second + 1);
        for (auto it : segs) add(it.first, it.second);
      }
    }
    reverse(ans.begin(), ans.end());
    for (auto it : ans) {
      printf("%d\n", it);
    }
    return 0;
  }
  