#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0;
  bool f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return f ? -x : x;
}
const int L = 1e7 + 10;
const int N = 1e5 + 10;
int n, m, val[L], ndn, ch[L][2], op[N], fa[L];
long long x[N], y[N];
bool ans[N];
vector<pair<int, int> > G[N];
long long tot;
inline int newd(int v) { return val[++ndn] = v, ndn; }
inline void pd(int x) {
  ch[x][0] = ch[x][0] ? ch[x][0] : newd(val[x]);
  ch[x][1] = ch[x][1] ? ch[x][1] : newd(val[x]);
  val[x] = 0;
}
void insert(int o, long long xl, long long xr, int v, long long l = 0,
            long long r = tot) {
  if (xl == l && xr == r) return val[o] = v, void();
  pd(o);
  long long mid = l + r >> 1;
  if (xr <= mid)
    insert(ch[o][0], xl, xr, v, l, mid);
  else if (xl > mid)
    insert(ch[o][1], xl, xr, v, mid + 1, r);
  else
    insert(ch[o][0], xl, mid, v, l, mid),
        insert(ch[o][1], mid + 1, xr, v, mid + 1, r);
}
void solve(int x, int y) {
  if (val[x] && val[y]) return G[min(val[x], val[y])].push_back({x, y}), void();
  if (val[x]) return solve(x, ch[y][0]), solve(x, ch[y][1]), void();
  if (val[y]) return solve(ch[x][0], y), solve(ch[x][1], y), void();
  return solve(ch[x][0], ch[y][0]), solve(ch[x][1], ch[y][1]);
}
int query(int o, long long pos, long long l = 0, long long r = tot) {
  if (val[o]) return o;
  long long mid = l + r >> 1;
  if (pos <= mid)
    return query(ch[o][0], pos, l, mid);
  else
    return query(ch[o][1], pos, mid + 1, r);
}
inline int gf(int x) { return fa[x] == x ? x : fa[x] = gf(fa[x]); }
inline void merge(int x, int y) {
  x = gf(x);
  y = gf(y);
  fa[x] = y;
}
int main() {
  n = read(), m = read();
  tot = 1ll << n;
  tot--;
  for (int i = (1); i <= (m); i++) {
    static char s[7];
    scanf("%s", s);
    if (s[0] == 'a')
      op[i] = 1;
    else
      op[i] = 0;
    x[i] = read(), y[i] = read();
  }
  newd(m + 1);
  for (int i = (1); i <= (m); i++)
    if (op[i] == 0) insert(1, x[i], y[i], i);
  for (int x = (1); x <= (ndn); x++)
    if (ch[x][0] && ch[x][1]) solve(ch[x][0], ch[x][1]);
  for (int i = (1); i <= (ndn); i++) fa[i] = i;
  for (int i = (m); i >= (1); i--) {
    for (auto d : G[i + 1]) merge(d.first, d.second);
    if (op[i]) {
      ans[i] = gf(query(1, x[i])) == gf(query(1, y[i]));
    }
  }
  for (int i = (1); i <= (m); i++)
    if (op[i]) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

<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;
long long read() {
  long long x = 0;
  bool f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return f ? -x : x;
}
const int L = 1e7 + 10;
const int N = 1e5 + 10;
int n, m, val[L], ndn, ch[L][2], op[N], fa[L];
long long x[N], y[N];
bool ans[N];
vector<pair<int, int> > G[N];
long long tot;
inline int newd(int v) { return val[++ndn] = v, ndn; }
inline void pd(int x) {
  ch[x][0] = ch[x][0] ? ch[x][0] : newd(val[x]);
  ch[x][1] = ch[x][1] ? ch[x][1] : newd(val[x]);
  val[x] = 0;
}
void insert(int o, long long xl, long long xr, int v, long long l = 0,
            long long r = tot) {
  if (xl == l && xr == r) return val[o] = v, void();
  pd(o);
  long long mid = l + r >> 1;
  if (xr <= mid)
    insert(ch[o][0], xl, xr, v, l, mid);
  else if (xl > mid)
    insert(ch[o][1], xl, xr, v, mid + 1, r);
  else
    insert(ch[o][0], xl, mid, v, l, mid),
        insert(ch[o][1], mid + 1, xr, v, mid + 1, r);
}
void solve(int x, int y) {
  if (val[x] && val[y]) return G[min(val[x], val[y])].push_back({x, y}), void();
  if (val[x]) return solve(x, ch[y][0]), solve(x, ch[y][1]), void();
  if (val[y]) return solve(ch[x][0], y), solve(ch[x][1], y), void();
  return solve(ch[x][0], ch[y][0]), solve(ch[x][1], ch[y][1]);
}
int query(int o, long long pos, long long l = 0, long long r = tot) {
  if (val[o]) return o;
  long long mid = l + r >> 1;
  if (pos <= mid)
    return query(ch[o][0], pos, l, mid);
  else
    return query(ch[o][1], pos, mid + 1, r);
}
inline int gf(int x) { return fa[x] == x ? x : fa[x] = gf(fa[x]); }
inline void merge(int x, int y) {
  x = gf(x);
  y = gf(y);
  fa[x] = y;
}
int main() {
  n = read(), m = read();
  tot = 1ll << n;
  tot--;
  for (int i = (1); i <= (m); i++) {
    static char s[7];
    scanf("%s", s);
    if (s[0] == 'a')
      op[i] = 1;
    else
      op[i] = 0;
    x[i] = read(), y[i] = read();
  }
  newd(m + 1);
  for (int i = (1); i <= (m); i++)
    if (op[i] == 0) insert(1, x[i], y[i], i);
  for (int x = (1); x <= (ndn); x++)
    if (ch[x][0] && ch[x][1]) solve(ch[x][0], ch[x][1]);
  for (int i = (1); i <= (ndn); i++) fa[i] = i;
  for (int i = (m); i >= (1); i--) {
    for (auto d : G[i + 1]) merge(d.first, d.second);
    if (op[i]) {
      ans[i] = gf(query(1, x[i])) == gf(query(1, y[i]));
    }
  }
  for (int i = (1); i <= (m); i++)
    if (op[i]) printf("%d\n", ans[i]);
  return 0;
}

```