#include <bits/stdc++.h>
using namespace std;
struct itv {
  unsigned long long l, r;
  itv(unsigned long long _l = 0, unsigned long long _r = 0) : l(_l), r(_r) {}
};
bool operator<(itv A, itv B) { return A.l < B.l; }
char op[10];
bool type[100000], ans[100000];
itv Q[100000];
vector<itv> E[2];
itv F[6000000];
int vcnt;
vector<int> ed[6000000];
bool ban[6000000];
void build(unsigned long long L, unsigned long long R, vector<itv> E) {
  if (E.size() == 1) {
    ++vcnt;
    F[vcnt] = itv(L, R);
    return;
  }
  unsigned long long mid = (L + R) >> 1;
  int pos = 0;
  while (pos + 1 < E.size() && E[pos + 1].l <= mid) pos++;
  int x = vcnt;
  vector<itv> G;
  for (int i = 0; i < pos; i++) G.push_back(E[i]);
  G.push_back(itv(E[pos].l, mid));
  build(L, mid, G);
  int y = vcnt;
  G.clear();
  if (E[pos].r > mid) G.push_back(itv(mid + 1, E[pos].r));
  for (int i = pos + 1; i < E.size(); i++) G.push_back(E[i]);
  build(mid + 1, R, G);
  int z = vcnt;
  unsigned long long C = R - mid;
  for (int i = x + 1, j = y + 1; i <= y && j <= z;) {
    ed[i].push_back(j), ed[j].push_back(i);
    if (F[i].r + C == F[j].r)
      i++, j++;
    else if (F[i].r + C < F[j].r)
      i++;
    else
      j++;
  }
}
struct BCJ {
  int fa[6000000];
  void init(int n) {
    for (int i = 1; i <= n; i++) fa[i] = i;
  }
  int fnd(int x) { return x == fa[x] ? x : fa[x] = fnd(fa[x]); }
  void conn(int x, int y) { fa[fnd(x)] = fnd(y); }
} B;
int main() {
  int n = 0, m = 0;
  scanf("%d%d", &n, &m);
  unsigned long long U = (1ull << n) - 1;
  for (int i = 1; i <= m; i++) {
    scanf("%s%llu%llu", op, &Q[i].l, &Q[i].r);
    if (op[0] == 'b') {
      type[i] = 0;
      E[0].push_back(Q[i]);
    } else
      type[i] = 1;
  }
  if (E[0].size() == 0) {
    for (int i = 1; i <= m; i++)
      if (type[i]) puts("1");
    return 0;
  }
  sort(E[0].begin(), E[0].end());
  if (E[0][0].l > 0) E[1].push_back(itv(0, E[0][0].l - 1));
  for (int i = 0; i < E[0].size() - 1; i++) {
    E[1].push_back(E[0][i]);
    if (E[0][i + 1].l > E[0][i].r + 1)
      E[1].push_back(itv(E[0][i].r + 1, E[0][i + 1].l - 1));
  }
  E[1].push_back(E[0].back());
  if (E[0].back().r < U) E[1].push_back(itv(E[0].back().r + 1, U));
  build(0, U, E[1]);
  for (int i = 0, j = 1; i < E[0].size(); i++) {
    while (j <= vcnt && F[j].r < E[0][i].l) j++;
    while (j <= vcnt && F[j].r <= E[0][i].r) {
      ban[j] = 1;
      j++;
    }
  }
  B.init(vcnt);
  for (int j = 1; j <= vcnt; j++)
    if (!ban[j]) {
      for (int k = 0; k < ed[j].size(); k++)
        if (!ban[ed[j][k]]) B.conn(j, ed[j][k]);
    }
  for (int i = m; i >= 1; i--) {
    if (type[i]) {
      int posx = upper_bound(F + 1, F + vcnt + 1, itv(Q[i].l, 0)) - F - 1,
          posy = upper_bound(F + 1, F + vcnt + 1, itv(Q[i].r, 0)) - F - 1;
      ans[i] = B.fnd(posx) == B.fnd(posy);
    } else {
      int posx = upper_bound(F + 1, F + vcnt + 1, itv(Q[i].l, 0)) - F - 1;
      while (posx <= vcnt && F[posx].r <= Q[i].r) {
        ban[posx] = 0;
        for (int j = 0; j < ed[posx].size(); j++)
          if (!ban[ed[posx][j]]) B.conn(ed[posx][j], posx);
        posx++;
      }
    }
  }
  for (int i = 1; i <= m; i++)
    if (type[i]) printf("%d\n", (int)ans[i]);
}

### Prompt

Your challenge is to write 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;
struct itv {
  unsigned long long l, r;
  itv(unsigned long long _l = 0, unsigned long long _r = 0) : l(_l), r(_r) {}
};
bool operator<(itv A, itv B) { return A.l < B.l; }
char op[10];
bool type[100000], ans[100000];
itv Q[100000];
vector<itv> E[2];
itv F[6000000];
int vcnt;
vector<int> ed[6000000];
bool ban[6000000];
void build(unsigned long long L, unsigned long long R, vector<itv> E) {
  if (E.size() == 1) {
    ++vcnt;
    F[vcnt] = itv(L, R);
    return;
  }
  unsigned long long mid = (L + R) >> 1;
  int pos = 0;
  while (pos + 1 < E.size() && E[pos + 1].l <= mid) pos++;
  int x = vcnt;
  vector<itv> G;
  for (int i = 0; i < pos; i++) G.push_back(E[i]);
  G.push_back(itv(E[pos].l, mid));
  build(L, mid, G);
  int y = vcnt;
  G.clear();
  if (E[pos].r > mid) G.push_back(itv(mid + 1, E[pos].r));
  for (int i = pos + 1; i < E.size(); i++) G.push_back(E[i]);
  build(mid + 1, R, G);
  int z = vcnt;
  unsigned long long C = R - mid;
  for (int i = x + 1, j = y + 1; i <= y && j <= z;) {
    ed[i].push_back(j), ed[j].push_back(i);
    if (F[i].r + C == F[j].r)
      i++, j++;
    else if (F[i].r + C < F[j].r)
      i++;
    else
      j++;
  }
}
struct BCJ {
  int fa[6000000];
  void init(int n) {
    for (int i = 1; i <= n; i++) fa[i] = i;
  }
  int fnd(int x) { return x == fa[x] ? x : fa[x] = fnd(fa[x]); }
  void conn(int x, int y) { fa[fnd(x)] = fnd(y); }
} B;
int main() {
  int n = 0, m = 0;
  scanf("%d%d", &n, &m);
  unsigned long long U = (1ull << n) - 1;
  for (int i = 1; i <= m; i++) {
    scanf("%s%llu%llu", op, &Q[i].l, &Q[i].r);
    if (op[0] == 'b') {
      type[i] = 0;
      E[0].push_back(Q[i]);
    } else
      type[i] = 1;
  }
  if (E[0].size() == 0) {
    for (int i = 1; i <= m; i++)
      if (type[i]) puts("1");
    return 0;
  }
  sort(E[0].begin(), E[0].end());
  if (E[0][0].l > 0) E[1].push_back(itv(0, E[0][0].l - 1));
  for (int i = 0; i < E[0].size() - 1; i++) {
    E[1].push_back(E[0][i]);
    if (E[0][i + 1].l > E[0][i].r + 1)
      E[1].push_back(itv(E[0][i].r + 1, E[0][i + 1].l - 1));
  }
  E[1].push_back(E[0].back());
  if (E[0].back().r < U) E[1].push_back(itv(E[0].back().r + 1, U));
  build(0, U, E[1]);
  for (int i = 0, j = 1; i < E[0].size(); i++) {
    while (j <= vcnt && F[j].r < E[0][i].l) j++;
    while (j <= vcnt && F[j].r <= E[0][i].r) {
      ban[j] = 1;
      j++;
    }
  }
  B.init(vcnt);
  for (int j = 1; j <= vcnt; j++)
    if (!ban[j]) {
      for (int k = 0; k < ed[j].size(); k++)
        if (!ban[ed[j][k]]) B.conn(j, ed[j][k]);
    }
  for (int i = m; i >= 1; i--) {
    if (type[i]) {
      int posx = upper_bound(F + 1, F + vcnt + 1, itv(Q[i].l, 0)) - F - 1,
          posy = upper_bound(F + 1, F + vcnt + 1, itv(Q[i].r, 0)) - F - 1;
      ans[i] = B.fnd(posx) == B.fnd(posy);
    } else {
      int posx = upper_bound(F + 1, F + vcnt + 1, itv(Q[i].l, 0)) - F - 1;
      while (posx <= vcnt && F[posx].r <= Q[i].r) {
        ban[posx] = 0;
        for (int j = 0; j < ed[posx].size(); j++)
          if (!ban[ed[posx][j]]) B.conn(ed[posx][j], posx);
        posx++;
      }
    }
  }
  for (int i = 1; i <= m; i++)
    if (type[i]) printf("%d\n", (int)ans[i]);
}

```