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 MAXN = 5e4 + 5, MAXM = 1e7;
  int n, m;
  long long a[MAXN], b[MAXN];
  int ans[MAXN];
  int cntp, rt, lc[MAXM], rc[MAXM], tag[MAXM];
  inline void add(int &k, int v) {
    if (!k) k = ++cntp;
    tag[k] = v;
  }
  inline void pushdwn(int k) {
    if (tag[k]) {
      add(lc[k], tag[k]);
      add(rc[k], tag[k]);
      tag[k] = 0;
    }
  }
  void Modify(int &k, long long l, long long r, long long x, long long y, int v) {
    if (!k) k = ++cntp;
    if (x <= l && r <= y) return add(k, v);
    pushdwn(k);
    long long mid = l + r >> 1;
    if (x <= mid) Modify(lc[k], l, mid, x, y, v);
    if (mid < y) Modify(rc[k], mid + 1, r, x, y, v);
    return;
  }
  int Query(int k, long long l, long long r, long long w) {
    if (tag[k]) return k;
    long long mid = l + r >> 1;
    if (w <= mid) return Query(lc[k], l, mid, w);
    return Query(rc[k], mid + 1, r, w);
  }
  int pre[MAXM];
  int fnd(int x) {
    if (x != pre[x]) pre[x] = fnd(pre[x]);
    return pre[x];
  }
  vector<pair<int, int> > vec[MAXN];
  void Dfs2(int k, int p) {
    if (tag[k] && tag[p]) {
      vec[min(tag[k], tag[p])].push_back(make_pair(k, p));
      return;
    }
    int l = lc[k], r = lc[p];
    if (!l) l = k;
    if (!r) r = p;
    Dfs2(l, r);
    l = rc[k], r = rc[p];
    if (!l) l = k;
    if (!r) r = p;
    Dfs2(l, r);
    return;
  }
  void Dfs1(int k) {
    if (!lc[k]) return;
    Dfs2(lc[k], rc[k]);
    Dfs1(lc[k]);
    Dfs1(rc[k]);
    return;
  }
  int main() {
    scanf("%d%d", &n, &m);
    add(rt, m + 1);
    for (int i = 1; i <= m; i++) {
      char op[10];
      scanf("%s%lld%lld", op, a + i, b + i);
      if (op[0] == 'b') ans[i] = -1, Modify(rt, 0, (1ll << n) - 1, a[i], b[i], i);
    }
    Dfs1(rt);
    for (int i = 1; i <= cntp; i++) pre[i] = i;
    ans[m + 1] = -1;
    for (int i = m + 1; i; i--) {
      for (int j = 0; j < vec[i].size(); j++) {
        int x = fnd(vec[i][j].first), y = fnd(vec[i][j].second);
        if (x != y) pre[x] = y;
      }
      if (!ans[i])
        ans[i] = (fnd(Query(rt, 0, (1ll << n) - 1, a[i])) ==
                  fnd(Query(rt, 0, (1ll << n) - 1, b[i])));
    }
    for (int i = 1; i <= m; i++)
      if (ans[i] >= 0) printf("%d\n", ans[i]);
    return 0;
  }
  