#include <bits/stdc++.h>
using namespace std;
const int M = 1e5 + 5, N = 105;
struct edge {
  int x, y;
};
vector<edge> e[M];
namespace SGT {
int lp[M * N], rp[M * N], tim[M * N], tot = 1;
inline bool leaf(int rt) { return !lp[rt] && !rp[rt]; }
inline void dn(int rt) {
  if (!lp[rt]) lp[rt] = ++tot;
  if (!rp[rt]) rp[rt] = ++tot;
  if (tim[rt]) tim[lp[rt]] = tim[rp[rt]] = tim[rt], tim[rt] = 0;
}
void update(int rt, long long l, long long r, long long ql, long long qr,
            int t) {
  if (ql <= l && r <= qr) return tim[rt] = t, void();
  dn(rt);
  long long mid = (l + r) >> 1;
  if (ql <= mid) update(lp[rt], l, mid, ql, qr, t);
  if (mid < qr) update(rp[rt], mid + 1, r, ql, qr, t);
}
void adde(int x, int y) {
  if (leaf(x) && leaf(y)) return e[min(tim[x], tim[y])].push_back({x, y});
  if (leaf(x)) return adde(x, lp[y]), adde(x, rp[y]);
  if (leaf(y)) return adde(lp[x], y), adde(rp[x], y);
  adde(lp[x], lp[y]), adde(rp[x], rp[y]);
}
int pos(int rt, long long l, long long r, long long p) {
  if (leaf(rt)) return rt;
  long long mid = (l + r) >> 1;
  return p <= mid ? pos(lp[rt], l, mid, p) : pos(rp[rt], mid + 1, r, p);
}
}  // namespace SGT
struct quy {
  bool opt;
  long long l, r;
} q[M];
int n, m, fa[M * N];
bool ans[M];
char ch[10];
int fd(int x) { return fa[x] == x ? x : fa[x] = fd(fa[x]); }
inline void mge(int x, int y) {
  x = fd(x), y = fd(y);
  fa[x] = y;
}
int main() {
  scanf("%d%d", &n, &m);
  long long mx = 1ll << n;
  SGT::tim[1] = m + 1;
  for (int i = 1; i <= m; ++i) {
    long long x, y;
    scanf("%s%lld%lld", ch, &x, &y);
    if (ch[0] == 'a')
      q[i] = {1, x, y};
    else
      SGT::update(1, 0, mx - 1, x, y, i), q[i].opt = 0;
  }
  for (int i = 1; i <= SGT::tot; ++i) {
    if (!SGT::leaf(i)) SGT::adde(SGT::lp[i], SGT::rp[i]);
    fa[i] = i;
  }
  for (auto c : e[m + 1]) mge(c.x, c.y);
  for (int i = m; i; --i) {
    for (auto c : e[i]) mge(c.x, c.y);
    if (q[i].opt)
      ans[i] = (fd(SGT::pos(1, 0, mx - 1, q[i].l)) ==
                fd(SGT::pos(1, 0, mx - 1, q[i].r)));
  }
  for (int i = 1; i <= m; ++i)
    if (q[i].opt) puts(ans[i] ? "1" : "0");
  return 0;
}

### Prompt

Generate 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;
const int M = 1e5 + 5, N = 105;
struct edge {
  int x, y;
};
vector<edge> e[M];
namespace SGT {
int lp[M * N], rp[M * N], tim[M * N], tot = 1;
inline bool leaf(int rt) { return !lp[rt] && !rp[rt]; }
inline void dn(int rt) {
  if (!lp[rt]) lp[rt] = ++tot;
  if (!rp[rt]) rp[rt] = ++tot;
  if (tim[rt]) tim[lp[rt]] = tim[rp[rt]] = tim[rt], tim[rt] = 0;
}
void update(int rt, long long l, long long r, long long ql, long long qr,
            int t) {
  if (ql <= l && r <= qr) return tim[rt] = t, void();
  dn(rt);
  long long mid = (l + r) >> 1;
  if (ql <= mid) update(lp[rt], l, mid, ql, qr, t);
  if (mid < qr) update(rp[rt], mid + 1, r, ql, qr, t);
}
void adde(int x, int y) {
  if (leaf(x) && leaf(y)) return e[min(tim[x], tim[y])].push_back({x, y});
  if (leaf(x)) return adde(x, lp[y]), adde(x, rp[y]);
  if (leaf(y)) return adde(lp[x], y), adde(rp[x], y);
  adde(lp[x], lp[y]), adde(rp[x], rp[y]);
}
int pos(int rt, long long l, long long r, long long p) {
  if (leaf(rt)) return rt;
  long long mid = (l + r) >> 1;
  return p <= mid ? pos(lp[rt], l, mid, p) : pos(rp[rt], mid + 1, r, p);
}
}  // namespace SGT
struct quy {
  bool opt;
  long long l, r;
} q[M];
int n, m, fa[M * N];
bool ans[M];
char ch[10];
int fd(int x) { return fa[x] == x ? x : fa[x] = fd(fa[x]); }
inline void mge(int x, int y) {
  x = fd(x), y = fd(y);
  fa[x] = y;
}
int main() {
  scanf("%d%d", &n, &m);
  long long mx = 1ll << n;
  SGT::tim[1] = m + 1;
  for (int i = 1; i <= m; ++i) {
    long long x, y;
    scanf("%s%lld%lld", ch, &x, &y);
    if (ch[0] == 'a')
      q[i] = {1, x, y};
    else
      SGT::update(1, 0, mx - 1, x, y, i), q[i].opt = 0;
  }
  for (int i = 1; i <= SGT::tot; ++i) {
    if (!SGT::leaf(i)) SGT::adde(SGT::lp[i], SGT::rp[i]);
    fa[i] = i;
  }
  for (auto c : e[m + 1]) mge(c.x, c.y);
  for (int i = m; i; --i) {
    for (auto c : e[i]) mge(c.x, c.y);
    if (q[i].opt)
      ans[i] = (fd(SGT::pos(1, 0, mx - 1, q[i].l)) ==
                fd(SGT::pos(1, 0, mx - 1, q[i].r)));
  }
  for (int i = 1; i <= m; ++i)
    if (q[i].opt) puts(ans[i] ? "1" : "0");
  return 0;
}

```