#include <bits/stdc++.h>
#define x first
#define y second
#define all(v) v.begin(), v.end()
#define chkmin(a, b) a = min(a, b)
#define chkmax(a, b) a = max(a, b)
using namespace std;
typedef long long ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef pair<ll, ll> pii;
typedef vector<pii> vii;
typedef vector<bool> vb;
/// You may use:
// The number of students
int n;
// The probability any given student is positive
double P;
// This function performs a test on a subset of samples.
// Its argument is a vector of Booleans of length N,
// where the i-th element is true if the i-th sample should be added to the mix.
// It returns true if (and only if) at least one of the samples in the mix is positive.
bool test_students(std::vector<bool> mask) {
assert(mask.size() == (size_t)n);
std::string mask_str(n, ' ');
for (int i = 0; i < n; i++)
mask_str[i] = mask[i] ? '1' : '0';
printf("Q %s\n", mask_str.c_str());
fflush(stdout);
char answer;
scanf(" %c", &answer);
return answer == 'P';
}
/// You should implement:
// This function will be called once for each test instance.
// It should use test_students to determine which samples are positive.
// It must return a vector of Booleans of length N,
// where the i-th element is true if and only if the i-th sample is positive.
mt19937 rng(time(0));
int mrand(int l, int r) {
return uniform_int_distribution<int>(l, r)(rng);
}
int FindNext(int l, int r, vi &ind) {
vb mask(n);
for (int i = l; i < r; i++) mask[ind[i]] = 1;
if (!test_students(mask)) return r;
for (int i = l; i < r; i++) mask[ind[i]] = 0;
int begin = l + 1, end = r;
while (begin < end) {
int mid = (begin + end) >> 1;
for (int i = l; i < mid; i++) mask[ind[i]] = 1;
if (test_students(mask)) end = mid;
else begin = mid + 1;
for (int i = l; i < mid; i++) mask[ind[i]] = 0;
}
return end - 1;
}
void Solve(int l, int r, vi &ind, vb &ans) {
while (l < r) {
int nx = FindNext(l, r, ind);
ans[nx] = 1;
l = nx + 1;
}
}
int cnt = 0;
void Solve2(int l, int r, vb &ans) {
if (l + 1 == r) {
cnt++;
ans[l] = 1;
return;
}
if (mrand(0, 1)) {
vb test(n);
int mid = (l + r) >> 1;
fill(test.begin() + l, test.begin() + mid, 1);
bool b = test_students(test);
fill(test.begin() + l, test.begin() + mid, 0);
if (!b) {
Solve2(mid, r, ans);
return;
}
Solve2(l, mid, ans);
if (cnt > 2 * n * P) return;
fill(test.begin() + mid, test.begin() + r, 1);
b = test_students(test);
fill(test.begin() + mid, test.begin() + r, 0);
if (b) Solve2(mid, r, ans);
} else {
vb test(n);
int mid = (l + r) >> 1;
fill(test.begin() + mid, test.begin() + r, 1);
bool b = test_students(test);
fill(test.begin() + mid, test.begin() + r, 0);
if (!b) {
Solve2(l, mid, ans);
return;
}
Solve2(mid, r, ans);
if (cnt > 2 * n * P) return;
fill(test.begin() + l, test.begin() + mid, 1);
b = test_students(test);
fill(test.begin() + l, test.begin() + mid, 0);
if (b) Solve2(l, mid, ans);
}
}
vb find_positive() {
if (P < 0.05) {
vb ans(n);
vi ind(n);
iota(all(ind), 0);
while (!ind.empty()) {
vi nx;
shuffle(all(ind), rng);
int m = ind.size();
int blocks = min<int>(m, ceil(m * P * 2));
int len = (m + blocks - 1) / blocks;
for (int i = 0; i < m; i += len) {
// Solve(i, min(i + len, n), ans);
int l = i, r = min(i + len, m);
int j = FindNext(l, r, ind);
if (j < r) ans[ind[j++]] = 1;
for (int k = j; k < r; k++) nx.push_back(ind[k]);
}
swap(ind, nx);
}
return ans;
}
vb ans(n);
cnt = 0;
Solve2(0, n, ans);
return ans;
// int block = 1000;
// for (int i = 0; i < n; i += block) Solve(i, i + block, ans);
// Solve(0, n, answer);
// vb test(n);
// for (int i = 0; i < n; i++) {
// test[i] = 1;
// answer[i] = test_students(test);
// test[i] = 0;
// }
// return ans;
}
int main() {
int T;
scanf("%d %lf %d", &n, &P, &T);
// You may perform any extra initialization here.
for (int i = 0; i < T; i++) {
vb answer = find_positive();
assert(answer.size() == (size_t)n);
string answer_str(n, ' ');
for (int j = 0; j < n; j++)
answer_str[j] = answer[j] ? '1' : '0';
printf("A %s\n", answer_str.c_str());
fflush(stdout);
char verdict;
scanf(" %c", &verdict);
if (verdict == 'W')
exit(0);
}
return 0;
}
