// Similar to Benq's solution, based on xor basis: https://oj.uz/submission/143064
// Also there is constructive solution like DeBrujin Sequences, see errorgorn's solution: https://oj.uz/submission/288552
#include <bits/stdc++.h>
using i64 = long long;
#ifdef DEBUG
#include "/home/ahmetalp/Desktop/Workplace/debug.h"
#else
#define debug(...) void(23)
#endif
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int N, K, T;
std::cin >> N >> K >> T;
std::string S;
std::cin >> S;
if (K % 2 == 0) {
std::cout << "-1\n";
return 0;
}
std::vector<std::pair<int, int>> basis;
for (int i = 0; i < (1 << N); ++i) {
if (__builtin_popcount(i) != K) {
continue;
}
int x = i;
for (auto[j, _] : basis) {
x = std::min(x, x ^ j);
}
if (x) {
basis.emplace_back(x, i);
}
}
debug(basis);
assert(basis.size() == N);
std::vector<int> ans(1 << N);
for (int i = 1; i < (1 << N); ++i) {
ans[i] = ans[i - 1] ^ basis[__builtin_ctz(i)].second;
}
debug(ans);
std::cout << (1 << N) << '\n';
for (int i = 0; i < (1 << N); ++i) {
std::string s(N, '?');
for (int j = 0; j < N; ++j) {
s[j] = char('0' + (ans[i] >> j & 1)) ^ (S[j] - '0');
}
std::cout << s << '\n';
}
return 0;
}
