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>
  #pragma GCC optimize(2)
  using namespace std;
  int n, m, fa[3000010], ch[3000010][2], id[3000010], cnt = 0, ndcnt = 0;
  bool vis[3000010];
  char buf[10];
  vector<pair<pair<long long, long long>, int> > qry;
  vector<long long> allv;
  vector<int> ans, vt[3000010];
  int getf(int x) { return x == fa[x] ? x : fa[x] = getf(fa[x]); }
  void add_seg(int cur, long long l, long long r, long long tl, long long tr) {
    if (l <= tl && tr <= r) {
      id[cur] = cnt++;
      return;
    }
    long long mid = (tl + tr) >> 1;
    if (mid >= l) {
      if (ch[cur][0] == -1) {
        ch[cur][0] = ++ndcnt;
      }
      add_seg(ch[cur][0], l, r, tl, mid);
    }
    if (mid < r) {
      if (ch[cur][1] == -1) {
        ch[cur][1] = ++ndcnt;
      }
      add_seg(ch[cur][1], l, r, mid + 1, tr);
    }
    return;
  }
  void dfs(int l, int r, int curdep, int dest) {
    if (id[l] != -1 && id[r] != -1) {
      if (curdep >= dest) {
        return;
      }
      vt[id[l]].push_back(id[r]);
      vt[id[r]].push_back(id[l]);
      return;
    }
    if (id[r] != -1) {
      swap(l, r);
    }
    if (id[l] != -1) {
      if (ch[r][0] != -1) {
        dfs(l, ch[r][0], curdep - 1, dest);
      }
      if (ch[r][1] != -1) {
        dfs(l, ch[r][1], curdep - 1, dest);
      }
      return;
    }
    if (curdep == dest) {
      if (l > r) {
        swap(l, r);
      }
      if (ch[l][0] != -1 && ch[r][1] != -1) {
        dfs(ch[l][0], ch[r][1], curdep - 1, dest);
      }
      return;
    }
    if (ch[l][0] != -1 && ch[r][0] != -1) {
      dfs(ch[l][0], ch[r][0], curdep - 1, dest);
    }
    if (ch[l][1] != -1 && ch[r][1] != -1) {
      dfs(ch[l][1], ch[r][1], curdep - 1, dest);
    }
    return;
  }
  int get_bel(long long x) {
    int cur = 0, i;
    for (i = n - 1; i >= 0; i--) {
      if (id[cur] != -1) {
        return id[cur];
      }
      if (x & ((long long)1 << i)) {
        cur = ch[cur][1];
      } else {
        cur = ch[cur][0];
      }
    }
    return id[cur];
  }
  void modify(long long l, long long r) {
    int i, j;
    l = get_bel(l);
    r = get_bel(r);
    for (i = l; i <= r; i++) {
      vis[i] = true;
      for (j = 0; j < vt[i].size(); j++) {
        if (vis[vt[i][j]]) {
          fa[getf(vt[i][j])] = getf(i);
        }
      }
    }
    return;
  }
  int main() {
    int i;
    long long l, r;
    for (i = 0; i < 3000010; i++) {
      id[i] = ch[i][0] = ch[i][1] = -1;
      fa[i] = i;
    }
    scanf("%d%d", &n, &m);
    for (i = 0; i < m; i++) {
      scanf("%s", &buf);
      scanf("%lld%lld", &l, &r);
      if (buf[0] == 'a') {
        qry.push_back(make_pair(make_pair(l, r), 0));
      } else {
        allv.push_back(l);
        allv.push_back(r);
        qry.push_back(make_pair(make_pair(l, r), 1));
      }
    }
    allv.push_back(-1);
    allv.push_back(((long long)1 << n));
    sort(allv.begin(), allv.end());
    long long rt = 0;
    for (i = 1; i < allv.size(); i++) {
      if (i & 1) {
        if (allv[i] - 1 >= allv[i - 1] + 1) {
          add_seg(0, allv[i - 1] + 1, allv[i] - 1, 0, ((long long)1 << n) - 1);
        }
      } else {
        add_seg(0, allv[i - 1], allv[i], 0, ((long long)1 << n) - 1);
      }
    }
    for (i = 0; i < n; i++) {
      dfs(0, 0, n - 1, i);
    }
    for (i = 1; i < allv.size(); i += 2) {
      if (allv[i] - 1 >= allv[i - 1] + 1) {
        modify(allv[i - 1] + 1, allv[i] - 1);
      }
    }
    reverse(qry.begin(), qry.end());
    for (i = 0; i < m; i++) {
      if (qry[i].second == 0) {
        ans.push_back(getf(get_bel(qry[i].first.first)) ==
                      getf(get_bel(qry[i].first.second)));
      } else {
        modify(qry[i].first.first, qry[i].first.second);
      }
    }
    reverse(ans.begin(), ans.end());
    for (i = 0; i < ans.size(); i++) {
      printf("%d\n", ans[i]);
    }
    return 0;
  }
  