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 long long N = 2e5 + 5, mod = 1e9 + 7;
  long long n, m, q, z, qt[N], ql[N], qr[N], ans[N], bl[N], br[N], asdf[N],
      f[N * 20], lg[N * 2], st[N][20], nid[N][20], sh[N][20];
  pair<long long, long long> a[N];
  vector<pair<long long, long long>> ed[N];
  long long find(long long u) { return f[u] == u ? u : f[u] = find(f[u]); }
  signed main() {
    scanf("%lld%lld", &n, &q);
    for (long long i = 2; i < 2 * N; i++) lg[i] = lg[i >> 1] + 1;
    a[++m] = {-1, -q - 1}, a[++m] = {1ll << n, q + 1};
    for (long long i = 1; i <= q; i++) {
      char s[10];
      scanf("%s%lld%lld", s, &ql[i], &qr[i]);
      qt[i] = (s[0] == 'b' ? 1 : 2);
      if (qt[i] == 1) a[++m] = {ql[i], -i}, a[++m] = {qr[i], i};
    }
    sort(a + 1, a + m + 1);
    for (long long i = 1; i < m; i++) {
      long long cl = a[i].first + (i & 1), cr = a[i + 1].first - (i & 1),
                cc = (i & 1 ? q + 1 : -a[i].second);
      if (cl > cr) continue;
      if (cl == cr)
        bl[++z] = cl, br[z] = cr, asdf[z] = cc;
      else {
        long long fk = cl ^ cr;
        while (fk != (fk & -fk)) fk ^= fk & -fk;
        long long mid = cl | (fk - 1);
        bl[++z] = cl, br[z] = mid, asdf[z] = cc, bl[++z] = mid + 1, br[z] = cr,
        asdf[z] = cc;
      }
    }
    long long ptr = z;
    for (long long i = 1; i <= z; i++)
      for (long long k = n - 1; k >= 0; k--) {
        long long cl, cr;
        bool die = false;
        if (br[i] - bl[i] > (1ll << k))
          cr = bl[i] | ((1ll << (k + 1)) - 1), cl = cr ^ ((1ll << (k + 1)) - 1),
          die = true;
        else
          cl = bl[i] ^ (1ll << k), cr = br[i] ^ (1ll << k);
        long long zl = lower_bound(br + 1, br + z + 1, cl) - br,
                  zr = lower_bound(bl + 1, bl + z + 1, cr + 1) - bl - 1;
        for (long long j = zl; j <= zr; j++)
          ed[min(asdf[j], asdf[i])].push_back({i, j});
        if (die) break;
      }
    for (long long i = 1; i <= ptr; i++) f[i] = i;
    for (long long i = q + 1; i >= 1; i--) {
      for (auto c : ed[i]) f[find(c.first)] = find(c.second);
      if (qt[i] == 2)
        ans[i] = find(lower_bound(br + 1, br + z + 1, ql[i]) - br) ==
                 find(lower_bound(br + 1, br + z + 1, qr[i]) - br);
    }
    for (long long i = 1; i <= q; i++)
      if (qt[i] == 2) printf("%lld\n", ans[i]);
    return 0;
  }
  