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 read(T &f) {
    f = 0;
    T fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') {
        fu = -1;
      }
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      f = (f << 3) + (f << 1) + (c & 15);
      c = getchar();
    }
    f *= fu;
  }
  template <typename T>
  void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x < 10)
      putchar(x + 48);
    else
      print(x / 10), putchar(x % 10 + 48);
  }
  template <typename T>
  void print(T x, char t) {
    print(x);
    putchar(t);
  }
  const int N = 2e5 + 5;
  vector<int> adj[N];
  long long a[N], b[N], l[N], r[N];
  int opt[N], id[N], ban[N], f[N], ans[N];
  char c[10];
  int n, m, len;
  int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
  bool cmp(int i, int j) { return l[i] < l[j]; }
  void add(long long L, long long R, int _ban) {
    for (int i = n; i >= 0; i--) {
      if (((R + 1) >> i) != (L >> i)) {
        long long pos = (((R + 1) >> i) << i) - 1;
        ++len;
        l[len] = L;
        r[len] = pos;
        ban[len] = _ban;
        if (pos != R) {
          ++len;
          l[len] = pos + 1;
          r[len] = R;
          ban[len] = _ban;
        }
        break;
      }
    }
  }
  void insert(int u) {
    for (int i = 0; i < (int)adj[u].size(); i++) {
      int v = adj[u][i];
      if (!ban[v]) {
        f[find(u)] = find(v);
      }
    }
  }
  int main() {
    read(n);
    read(m);
    for (int i = 1; i <= m; i++) {
      scanf("%s", c + 1);
      read(a[i]);
      read(b[i]);
      opt[i] = (c[1] == 'a');
      if (opt[i] == 0) {
        ++len;
        l[len] = a[i];
        r[len] = b[i];
      }
    }
    {
      static long long tl[N], tr[N], tlen;
      for (int i = 1; i <= len; i++) id[i] = i;
      sort(id + 1, id + len + 1, cmp);
      for (int i = 1; i <= len; i++) {
        tl[i] = l[id[i]];
        tr[i] = r[id[i]];
      }
      tlen = len;
      len = 0;
      tr[0] = -1;
      for (int i = 1; i <= tlen; i++) {
        if (tr[i - 1] + 1 <= tl[i] - 1) {
          add(tr[i - 1] + 1, tl[i] - 1, 0);
        }
        add(tl[i], tr[i], 1);
      }
      if (tr[tlen] != (1ll << n) - 1) add(tr[tlen] + 1, (1ll << n) - 1, 0);
    }
    for (int bit = 0; bit < n; bit++) {
      for (int i = 1, j = 1; i <= len; i++) {
        long long pos = -1;
        if ((l[i] >> bit) & 1)
          pos = l[i];
        else
          pos = ((l[i] >> bit) + 1) << bit;
        if (l[i] <= pos && pos <= r[i]) {
          while (pos - (1ll << bit) > r[j]) ++j;
          adj[i].push_back(j);
          adj[j].push_back(i);
        }
      }
      for (int i = 1, j = 1; i <= len; i++) {
        long long pos = -1;
        if (((r[i] >> bit) & 1) == 0)
          pos = l[i];
        else
          pos = (r[i] >> (bit + 1)) << (bit + 1);
        if (l[i] <= pos && pos <= r[i]) {
          while (pos + (1ll << bit) > r[j]) ++j;
          adj[i].push_back(j);
          adj[j].push_back(i);
        }
      }
    }
    for (int i = 1; i <= len; i++) f[i] = i;
    for (int i = 1; i <= len; i++) {
      if (!ban[i]) {
        insert(i);
      }
    }
    for (int i = m; i >= 1; i--) {
      if (opt[i] == 0) {
        int pos = lower_bound(r + 1, r + len + 1, b[i]) - r;
        while (pos && l[pos] >= a[i]) {
          ban[pos] = 0;
          insert(pos);
          --pos;
        }
      }
      if (opt[i] == 1) {
        int p1 = lower_bound(r + 1, r + len + 1, a[i]) - r;
        int p2 = lower_bound(r + 1, r + len + 1, b[i]) - r;
        if (find(p1) == find(p2)) ans[i] = 1;
      }
    }
    for (int i = 1; i <= m; i++) {
      if (opt[i] == 1) {
        print(ans[i], '\n');
      }
    }
    return 0;
  }
  