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>
  template <typename T>
  inline void read(T &x) {
    x = 0;
    char ch = getchar();
    while (!isdigit(ch)) ch = getchar();
    while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
  }
  using namespace std;
  const int N = 55, M = 200005;
  struct DSU {
    int par[M];
    inline void init(int n) { iota(par + 1, par + n + 1, 1); }
    int get(int x) {
      if (x == par[x]) return x;
      return par[x] = get(par[x]);
    }
    inline bool test(int x, int y) { return get(x) == get(y); }
    inline void merge(int x, int y) { par[get(x)] = get(y); }
  } dsu;
  int n, m, t[M], cnt, tot, tim[M];
  long long a[M], b[M], p[M], ptr[M];
  vector<int> g[M];
  int main() {
    read(n), read(m);
    for (int i = 1; i <= m; ++i) {
      static char s[10];
      scanf("%s", s);
      if (s[0] == 'b') {
        read(a[i]), read(b[i]);
        p[++cnt] = a[i], p[++cnt] = ++b[i];
      } else {
        read(a[i]), read(b[i]);
        t[i] = 1;
      }
    }
    p[++cnt] = 0, p[++cnt] = 1ll << n;
    sort(p + 1, p + cnt + 1);
    cnt = unique(p + 1, p + cnt + 1) - p - 1;
    ptr[++tot] = 0;
    for (int i = 1; i < cnt; ++i) {
      long long l = p[i], r = p[i + 1] - 1;
      if (l == r) {
        ptr[++tot] = l;
      } else {
        for (int j = n - 1; ~j; --j) {
          if (!((l ^ r) >> j & 1)) continue;
          ptr[++tot] = l;
          r &= ((1ll << (n - j)) - 1) << j;
          ptr[++tot] = r;
          break;
        }
      }
    }
    ptr[++tot] = 1ll << n;
    tot = unique(ptr + 1, ptr + tot + 1) - ptr - 1;
    auto judge = [&](int i, int j, int k) {
      long long l1 = ptr[i], r1 = ptr[i + 1] - 1;
      long long l2 = ptr[j], r2 = ptr[j + 1] - 1;
      return (l2 - r1 <= (1ll << k)) && (r2 - l1 >= (1ll << k));
    };
    auto add_edge = [&](int i, int j, int k) {
      long long l1 = ptr[i], r1 = ptr[i + 1] - 1;
      long long l2 = ptr[j], r2 = ptr[j + 1] - 1;
      if (l1 >> k & 1)
        l1 += (!k) ? 1 : ((1ll << (k - 1)) - (l1 & ((1ll << k) - 1)));
      if (r1 >> k & 1) r1 -= (!k) ? 1 : ((r1 & ((1ll << k) - 1)) + 1);
      if (l1 > r1) return;
      if (!(l2 >> k & 1))
        l2 += (!k) ? 1 : ((1ll << (k - 1)) - (l2 & ((1ll << k) - 1)));
      if (!(r2 >> k & 1)) r2 -= (!k) ? 1 : ((r2 & ((1ll << k) - 1)) + 1);
      if (l2 > r2) return;
      if ((l2 - r1 <= (1ll << k)) && (r2 - l1 >= (1ll << k))) {
        if (tim[i] > tim[j]) swap(i, j);
        g[i].emplace_back(j);
      }
    };
    for (int i = 1; i < tot; ++i) tim[i] = m + 1;
    for (int i = 1; i <= m; ++i) {
      if (t[i]) continue;
      int pos = lower_bound(ptr + 1, ptr + tot + 1, a[i]) - ptr;
      assert(ptr[pos] == a[i]);
      tim[pos] = i;
      if (a[i] + 1 != b[i]) tim[pos + 1] = i;
    }
    for (int k = 0; k < n; ++k) {
      for (int i = 1, j = 0; i < tot; ++i) {
        while (j && judge(j, i, k)) --j;
        for (; j + 1 < i && judge(j + 1, i, k); add_edge(++j, i, k))
          ;
      }
    }
    dsu.init(tot - 1);
    for (int i = 1; i < tot; ++i)
      if (tim[i] == m + 1)
        for (auto j : g[i]) dsu.merge(i, j);
    vector<int> ans;
    for (int i = m; i; --i) {
      if (t[i]) {
        int pa = upper_bound(ptr + 1, ptr + tot + 1, a[i]) - ptr - 1;
        int pb = upper_bound(ptr + 1, ptr + tot + 1, b[i]) - ptr - 1;
        ans.emplace_back(dsu.test(pa, pb));
      } else {
        int u = lower_bound(ptr + 1, ptr + tot + 1, a[i]) - ptr;
        for (auto v : g[u]) dsu.merge(u, v);
        if (a[i] + 1 != b[i]) {
          ++u;
          for (auto v : g[u]) dsu.merge(u, v);
        }
      }
    }
    reverse(ans.begin(), ans.end());
    for (auto x : ans) printf("%d\n", x);
    return 0;
  }
  