#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 2e5 + 555;
int n, m, q, z;
pair<long long, int> a[N];
int qt[N];
long long ql[N], qr[N];
long long ans[N];
vector<pair<int, int> > ed[N];
long long bl[N], br[N], arin[N];
int f[N * 20];
const int iu = 18;
int lg[N * 2];
long long st[N][iu], nid[N][iu];
long long sh[N][iu];
void adde(int u, int v, int t) { ed[t].push_back({u, v}); }
void add(int id, int l, int r) {
  for (int i = l; i <= r; i++) {
    adde(id, i, min(arin[i], arin[id]));
  }
}
int find(int x) {
  if (f[x] != x) f[x] = find(f[x]);
  return f[x];
}
void join(int x, int y) {
  x = find(x);
  y = find(y);
  f[x] = y;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 2; i < 262144; i++) lg[i] = lg[i / 2] + 1;
  a[++m] = {-1, -q - 1};
  a[++m] = {1LL << n, q + 1};
  for (int i = 1; i <= q; i++) {
    string s;
    cin >> s >> ql[i] >> qr[i];
    if (s[0] == 'b')
      qt[i] = 1;
    else
      qt[i] = 2;
    if (qt[i] == 1) {
      a[++m] = {ql[i], -i};
      a[++m] = {qr[i], i};
    }
  }
  sort(a + 1, a + m + 1);
  for (int i = 1; i < m; i++) {
    long long cl = a[i].first + (i % 2);
    long long cr = a[i + 1].first - (i % 2);
    int cc;
    if (i % 2 == 1)
      cc = q + 1;
    else
      cc = -a[i].second;
    if (cl > cr) continue;
    if (cl == cr) {
      bl[++z] = cl;
      br[z] = cr;
      arin[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;
      arin[z] = cc;
      bl[++z] = mid + 1;
      br[z] = cr;
      arin[z] = cc;
    }
  }
  int ptr = z;
  for (int i = 1; i <= z; i++) {
    for (int k = n - 1; k >= 0; k--) {
      long long cl, cr;
      bool die = false;
      if (br[i] - bl[i] + 1 > (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);
      }
      int zl = lower_bound(br + 1, br + z + 1, cl) - br;
      int zr = lower_bound(bl + 1, bl + z + 1, cr + 1) - bl - 1;
      add(i, zl, zr);
      if (die) break;
    }
  }
  for (int i = 1; i <= ptr; i++) f[i] = i;
  for (int i = q + 1; i >= 1; i--) {
    for (auto c : ed[i]) join(c.first, c.second);
    if (qt[i] == 2) {
      int cl = lower_bound(br + 1, br + z + 1, ql[i]) - br;
      int cr = lower_bound(br + 1, br + z + 1, qr[i]) - br;
      ans[i] = find(cl) == find(cr);
    }
  }
  int out = 0;
  for (int i = 1; i <= q; i++) {
    if (qt[i] == 2) {
      cout << ans[i] << '\n';
    }
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

<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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 2e5 + 555;
int n, m, q, z;
pair<long long, int> a[N];
int qt[N];
long long ql[N], qr[N];
long long ans[N];
vector<pair<int, int> > ed[N];
long long bl[N], br[N], arin[N];
int f[N * 20];
const int iu = 18;
int lg[N * 2];
long long st[N][iu], nid[N][iu];
long long sh[N][iu];
void adde(int u, int v, int t) { ed[t].push_back({u, v}); }
void add(int id, int l, int r) {
  for (int i = l; i <= r; i++) {
    adde(id, i, min(arin[i], arin[id]));
  }
}
int find(int x) {
  if (f[x] != x) f[x] = find(f[x]);
  return f[x];
}
void join(int x, int y) {
  x = find(x);
  y = find(y);
  f[x] = y;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 2; i < 262144; i++) lg[i] = lg[i / 2] + 1;
  a[++m] = {-1, -q - 1};
  a[++m] = {1LL << n, q + 1};
  for (int i = 1; i <= q; i++) {
    string s;
    cin >> s >> ql[i] >> qr[i];
    if (s[0] == 'b')
      qt[i] = 1;
    else
      qt[i] = 2;
    if (qt[i] == 1) {
      a[++m] = {ql[i], -i};
      a[++m] = {qr[i], i};
    }
  }
  sort(a + 1, a + m + 1);
  for (int i = 1; i < m; i++) {
    long long cl = a[i].first + (i % 2);
    long long cr = a[i + 1].first - (i % 2);
    int cc;
    if (i % 2 == 1)
      cc = q + 1;
    else
      cc = -a[i].second;
    if (cl > cr) continue;
    if (cl == cr) {
      bl[++z] = cl;
      br[z] = cr;
      arin[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;
      arin[z] = cc;
      bl[++z] = mid + 1;
      br[z] = cr;
      arin[z] = cc;
    }
  }
  int ptr = z;
  for (int i = 1; i <= z; i++) {
    for (int k = n - 1; k >= 0; k--) {
      long long cl, cr;
      bool die = false;
      if (br[i] - bl[i] + 1 > (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);
      }
      int zl = lower_bound(br + 1, br + z + 1, cl) - br;
      int zr = lower_bound(bl + 1, bl + z + 1, cr + 1) - bl - 1;
      add(i, zl, zr);
      if (die) break;
    }
  }
  for (int i = 1; i <= ptr; i++) f[i] = i;
  for (int i = q + 1; i >= 1; i--) {
    for (auto c : ed[i]) join(c.first, c.second);
    if (qt[i] == 2) {
      int cl = lower_bound(br + 1, br + z + 1, ql[i]) - br;
      int cr = lower_bound(br + 1, br + z + 1, qr[i]) - br;
      ans[i] = find(cl) == find(cr);
    }
  }
  int out = 0;
  for (int i = 1; i <= q; i++) {
    if (qt[i] == 2) {
      cout << ans[i] << '\n';
    }
  }
}

```