#include <bits/stdc++.h>
using namespace std;
int n, m, k = 0;
long long b[200010];
int fa[200010], ans[50010];
vector<int> E[200010];
bool ban[200010];
char s[10];
struct ask {
  int ty;
  long long l, r;
} q[50010];
int find(int u) { return u == fa[u] ? u : fa[u] = find(fa[u]); }
void merge(int u, int v) { u = find(u), v = find(v), fa[u] = v; }
void ins(int u) {
  for (int v : E[u])
    if (!ban[v]) merge(u, v);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &n);
  for (int i = 1; i <= m; i++) {
    scanf("%s%lld%lld", s + 1, &q[i].l, &q[i].r);
    q[i].ty = s[1] == 'a';
    if (!q[i].ty) b[++k] = q[i].l, b[++k] = q[i].r + 1;
  }
  b[++k] = 1ll << n, b[++k] = 0;
  sort(b + 1, b + k + 1);
  int o = k;
  for (int i = 1; i < o; i++) {
    long long l = b[i], r = b[i + 1] - 1,
              x = 1ll << (63 - __builtin_clzll(l ^ r));
    if (l >= r) continue;
    b[++k] = ((l / x) * x) | x;
  }
  sort(b + 1, b + k + 1);
  k = unique(b + 1, b + k + 1) - b - 2;
  for (int i = 0; i < n; i++)
    for (int p = 0; p < 2; p++)
      for (int j = 1, l = 1; j <= k; j++) {
        if (((b[j] >> i) & 1) == p) {
          while ((b[j] ^ (1ll << i)) >= b[l + 1]) l++;
          if (l != j) E[l].push_back(j), E[j].push_back(l);
        }
      }
  for (int i = 1; i <= m; i++) {
    q[i].l = upper_bound(b + 1, b + k + 1, q[i].l) - b - 1;
    q[i].r = upper_bound(b + 1, b + k + 1, q[i].r) - b - 1;
    if (!q[i].ty)
      for (int j = q[i].l; j <= q[i].r; j++) ban[j] = 1;
  }
  iota(fa + 1, fa + k + 1, 1);
  for (int i = 1; i <= k; i++)
    if (!ban[i]) ins(i);
  for (int i = m; i >= 1; i--) {
    if (!q[i].ty)
      for (int j = q[i].l; j <= q[i].r; j++) ban[j] = 0, ins(j);
    else
      ans[i] = find(q[i].l) == find(q[i].r);
  }
  for (int i = 1; i <= m; i++)
    if (q[i].ty) printf("%d\n", ans[i]);
}

### Prompt

In CPP, your task is to solve 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;
int n, m, k = 0;
long long b[200010];
int fa[200010], ans[50010];
vector<int> E[200010];
bool ban[200010];
char s[10];
struct ask {
  int ty;
  long long l, r;
} q[50010];
int find(int u) { return u == fa[u] ? u : fa[u] = find(fa[u]); }
void merge(int u, int v) { u = find(u), v = find(v), fa[u] = v; }
void ins(int u) {
  for (int v : E[u])
    if (!ban[v]) merge(u, v);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &n);
  for (int i = 1; i <= m; i++) {
    scanf("%s%lld%lld", s + 1, &q[i].l, &q[i].r);
    q[i].ty = s[1] == 'a';
    if (!q[i].ty) b[++k] = q[i].l, b[++k] = q[i].r + 1;
  }
  b[++k] = 1ll << n, b[++k] = 0;
  sort(b + 1, b + k + 1);
  int o = k;
  for (int i = 1; i < o; i++) {
    long long l = b[i], r = b[i + 1] - 1,
              x = 1ll << (63 - __builtin_clzll(l ^ r));
    if (l >= r) continue;
    b[++k] = ((l / x) * x) | x;
  }
  sort(b + 1, b + k + 1);
  k = unique(b + 1, b + k + 1) - b - 2;
  for (int i = 0; i < n; i++)
    for (int p = 0; p < 2; p++)
      for (int j = 1, l = 1; j <= k; j++) {
        if (((b[j] >> i) & 1) == p) {
          while ((b[j] ^ (1ll << i)) >= b[l + 1]) l++;
          if (l != j) E[l].push_back(j), E[j].push_back(l);
        }
      }
  for (int i = 1; i <= m; i++) {
    q[i].l = upper_bound(b + 1, b + k + 1, q[i].l) - b - 1;
    q[i].r = upper_bound(b + 1, b + k + 1, q[i].r) - b - 1;
    if (!q[i].ty)
      for (int j = q[i].l; j <= q[i].r; j++) ban[j] = 1;
  }
  iota(fa + 1, fa + k + 1, 1);
  for (int i = 1; i <= k; i++)
    if (!ban[i]) ins(i);
  for (int i = m; i >= 1; i--) {
    if (!q[i].ty)
      for (int j = q[i].l; j <= q[i].r; j++) ban[j] = 0, ins(j);
    else
      ans[i] = find(q[i].l) == find(q[i].r);
  }
  for (int i = 1; i <= m; i++)
    if (q[i].ty) printf("%d\n", ans[i]);
}

```