7 7 7
1
2
3
4
5
6
7
 4 2 8
1
3
4
2
3
4
1
2
 9 4 9
1
2
3
4
5
6
7
8
9
 7 5 7
1
2
3
4
5
6
7
 15 10 15
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
 4 0 4
1
2
3
4
 12 2 12
1
2
3
4
5
6
7
8
9
10
11
12
 10 2 10
1
2
3
4
5
6
7
8
9
10
 7 4 7
1
2
3
4
5
6
7
 7 3 7
1
2
3
4
5
6
7

XXXXXXX
 ..XX.X.X
 ...X.X.XX
 .X.XXXX
 .X.X.X.X.XXXXXX
 ....
 .........X.X
 .......X.X
 .X.X.XX
 ...X.XX

#include <bits/stdc++.h>
const long double PI = 3.141592653589793238L;
const long long N = 1000005;
using namespace std;
long long f(string x) {
  long long nb = 0;
  string s = x + x;
  for (long long i = (long long)(0);
       i <= (long long)(((long long)(s).size()) / 2 - 1); ++i) {
    long long j = i;
    while (s[j] != 'X') ++j;
    if ((j - i) & 1) ++nb;
  }
  return nb;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k, p;
  cin >> n >> k >> p;
  if (k == 0) {
    while (p--) cout << ".";
    return 0;
  }
  if (k == 1) {
    while (p--) {
      long long i;
      cin >> i;
      cout << ((i == n) ? "X" : ".");
    }
    return 0;
  }
  long long nb;
  if (2 * k >= n)
    nb = 2 * k - n;
  else if ((n - 2 * k) & 1)
    nb = 1;
  else
    nb = 0;
  while (p--) {
    long long i;
    cin >> i;
    if (i + nb > n)
      cout << "X";
    else if (i + (k - nb) * 2 + nb > n)
      cout << (((i & 1) == ((n - nb) & 1)) ? "X" : ".");
    else
      cout << ".";
  }
  return 0;
}
