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;
  int cl[5000010], cr[5000010], ld[5000010], sl = 1;
  bool yz[5000010];
  void pushd(int x) {
    if (cl[x] == 0) {
      cl[x] = ++sl;
      yz[cl[x]] = 1;
    }
    if (cr[x] == 0) {
      cr[x] = ++sl;
      yz[cr[x]] = 1;
    }
    yz[x] = 0;
    if (ld[x] == 0) return;
    ld[cl[x]] = ld[cr[x]] = ld[x];
    ld[x] = 0;
  }
  vector<int> cf[50010], ve[5000010];
  void addqj(int i, long long l, long long r, long long L, long long R, int x) {
    if (R <= l || r <= L) return;
    if (L <= l && r <= R) {
      ld[i] = x;
      cf[x].push_back(i);
      return;
    }
    long long m = (l + r) >> 1;
    pushd(i);
    addqj(cl[i], l, m, L, R, x);
    addqj(cr[i], m, r, L, R, x);
  }
  void link(int x, int y) {
    if (yz[x] && yz[y]) {
      if (ld[x] < ld[y])
        ve[x].push_back(y);
      else
        ve[y].push_back(x);
      return;
    }
    link(yz[x] ? x : cl[x], yz[y] ? y : cl[y]);
    link(yz[x] ? x : cr[x], yz[y] ? y : cr[y]);
  }
  int find(int i, long long l, long long r, long long x) {
    if (yz[i]) return i;
    long long m = (l + r) >> 1;
    return x < m ? find(cl[i], l, m, x) : find(cr[i], m, r, x);
  }
  int lx[50010], fa[5000010], ans[50010];
  long long A[50010], B[50010];
  int getv(int x) { return x == fa[x] ? x : fa[x] = getv(fa[x]); }
  void merge(int x, int y) {
    x = getv(x);
    y = getv(y);
    if (x != y) fa[x] = y;
  }
  int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    ld[1] = m + 1;
    yz[1] = 1;
    for (int i = 1; i <= m; i++) {
      char zf[10];
      scanf("%s%lld%lld", zf, &A[i], &B[i]);
      lx[i] = zf[0] == 'b' ? 0 : 1;
      if (lx[i] == 0) addqj(1, 0, (1ll << n), A[i], B[i] + 1, i);
    }
    for (int i = 1; i <= sl; i++) {
      fa[i] = i;
      if (!yz[i]) link(cl[i], cr[i]);
    }
    for (int i = 1; i <= sl; i++) {
      if (ld[i] != m + 1) continue;
      for (int j = 0; j < ve[i].size(); j++) merge(i, ve[i][j]);
    }
    for (int i = m; i >= 1; i--) {
      if (lx[i])
        ans[i] = (getv(find(1, 0, (1ll << n), A[i])) ==
                  getv(find(1, 0, (1ll << n), B[i])));
      else {
        for (int x = 0; x < cf[i].size(); x++) {
          int u = cf[i][x];
          for (int y = 0; y < ve[u].size(); y++) merge(u, ve[u][y]);
        }
      }
    }
    for (int i = 1; i <= m; i++) {
      if (lx[i]) printf("%d\n", ans[i]);
    }
    return 0;
  }
  