#include <bits/stdc++.h>
long long n, m;
long long que[50001][3], L[200010], top;
std::vector<long long> con[200010];
bool vis[200010], vis2[200010];
char str[11];
long long f(long long x) {
  if (!(x & (x - 1))) return x;
  return f(x & (x - 1));
}
long long fa[200010];
long long find(long long x) { return fa[x] ? fa[x] = find(fa[x]) : x; }
void add(long long x) {
  vis2[x] = 1;
  for (long long i = 0; i < con[x].size(); ++i) {
    if (!vis2[con[x][i]]) continue;
    long long u = find(x), v = find(con[x][i]);
    if (u != v) fa[u] = v;
  }
}
long long ans[50001], at;
signed main() {
  scanf("%lld%lld", &n, &m);
  L[++top] = 0;
  for (long long i = 1; i <= m; ++i) {
    scanf("%s%lld%lld", str + 1, &que[i][1], &que[i][2]);
    que[i][0] = str[1] == 'b';
    if (que[i][0]) {
      L[++top] = que[i][1];
      L[++top] = que[i][2] + 1;
    }
  }
  std::sort(L + 1, L + top + 1);
  long long now = top;
  L[++top] = 1ll << n;
  for (long long i = 1; i <= now; ++i) {
    long long l = L[i], r = L[i + 1] - 1, x = f(l ^ r);
    if (l == r || (!(l & (x - 1)))) continue;
    L[++top] = (l ^ (l & (x - 1))) | x;
  }
  std::sort(L + 1, L + top + 1);
  --top;
  for (long long i = 0; i < n; ++i)
    for (long long j = 0; j < 2; ++j) {
      for (long long k = 1, now = 1; k <= top; ++k) {
        long long l = L[k], r = L[k + 1] - 1, v;
        if (((l >> i) & 1) == j)
          v = l;
        else
          v = ((l >> i) + 1) << i;
        if (v <= r) {
          v ^= 1ll << i;
          while (v >= L[now + 1]) ++now;
          con[k].push_back(now);
          con[now].push_back(k);
        }
      }
    }
  for (long long i = m; i; --i)
    if (que[i][0]) {
      long long l = que[i][1], r = que[i][2] + 1;
      long long p = std::lower_bound(L + 1, L + top + 1, r) - L - 1;
      for (; p && L[p] >= l; --p) vis[p] = 1;
    }
  for (long long i = 1; i <= top; ++i)
    if (!vis[i]) add(i);
  for (long long i = m; i; --i)
    if (que[i][0]) {
      long long l = que[i][1], r = que[i][2] + 1;
      long long p = std::lower_bound(L + 1, L + top + 1, r) - L - 1;
      for (; p && L[p] >= l; --p) add(p);
    } else {
      long long a = std::upper_bound(L + 1, L + top + 1, que[i][1]) - L - 1,
                b = std::upper_bound(L + 1, L + top + 1, que[i][2]) - L - 1;
      ans[++at] = find(a) == find(b);
    }
  for (long long i = at; i; --i) printf("%d\n", ans[i]);
}

### 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>
long long n, m;
long long que[50001][3], L[200010], top;
std::vector<long long> con[200010];
bool vis[200010], vis2[200010];
char str[11];
long long f(long long x) {
  if (!(x & (x - 1))) return x;
  return f(x & (x - 1));
}
long long fa[200010];
long long find(long long x) { return fa[x] ? fa[x] = find(fa[x]) : x; }
void add(long long x) {
  vis2[x] = 1;
  for (long long i = 0; i < con[x].size(); ++i) {
    if (!vis2[con[x][i]]) continue;
    long long u = find(x), v = find(con[x][i]);
    if (u != v) fa[u] = v;
  }
}
long long ans[50001], at;
signed main() {
  scanf("%lld%lld", &n, &m);
  L[++top] = 0;
  for (long long i = 1; i <= m; ++i) {
    scanf("%s%lld%lld", str + 1, &que[i][1], &que[i][2]);
    que[i][0] = str[1] == 'b';
    if (que[i][0]) {
      L[++top] = que[i][1];
      L[++top] = que[i][2] + 1;
    }
  }
  std::sort(L + 1, L + top + 1);
  long long now = top;
  L[++top] = 1ll << n;
  for (long long i = 1; i <= now; ++i) {
    long long l = L[i], r = L[i + 1] - 1, x = f(l ^ r);
    if (l == r || (!(l & (x - 1)))) continue;
    L[++top] = (l ^ (l & (x - 1))) | x;
  }
  std::sort(L + 1, L + top + 1);
  --top;
  for (long long i = 0; i < n; ++i)
    for (long long j = 0; j < 2; ++j) {
      for (long long k = 1, now = 1; k <= top; ++k) {
        long long l = L[k], r = L[k + 1] - 1, v;
        if (((l >> i) & 1) == j)
          v = l;
        else
          v = ((l >> i) + 1) << i;
        if (v <= r) {
          v ^= 1ll << i;
          while (v >= L[now + 1]) ++now;
          con[k].push_back(now);
          con[now].push_back(k);
        }
      }
    }
  for (long long i = m; i; --i)
    if (que[i][0]) {
      long long l = que[i][1], r = que[i][2] + 1;
      long long p = std::lower_bound(L + 1, L + top + 1, r) - L - 1;
      for (; p && L[p] >= l; --p) vis[p] = 1;
    }
  for (long long i = 1; i <= top; ++i)
    if (!vis[i]) add(i);
  for (long long i = m; i; --i)
    if (que[i][0]) {
      long long l = que[i][1], r = que[i][2] + 1;
      long long p = std::lower_bound(L + 1, L + top + 1, r) - L - 1;
      for (; p && L[p] >= l; --p) add(p);
    } else {
      long long a = std::upper_bound(L + 1, L + top + 1, que[i][1]) - L - 1,
                b = std::upper_bound(L + 1, L + top + 1, que[i][2]) - L - 1;
      ans[++at] = find(a) == find(b);
    }
  for (long long i = at; i; --i) printf("%d\n", ans[i]);
}

```