#include #include #include #include #include #include #include #include #define rep(i, a, b) for (int i = int(a); i < int(b); i++) using namespace std; using ll = long long int; // NOLINT using P = pair; // clang-format off #ifdef _DEBUG_ #define dump(...) do{ cerr << __LINE__ << ":\t" << #__VA_ARGS__ << " = "; debug_print(__VA_ARGS__); } while(false) template void debug_print(const T &t, const Ts &...ts) { cerr << t; ((cerr << ", " << ts), ...); cerr << endl; } #else #define dump(...) do{ } while(false) #endif template vector make_v(size_t a, T b) { return vector(a, b); } template auto make_v(size_t a, Ts... ts) { return vector(a, make_v(ts...)); } template bool chmin(T &a, const T& b) { if (a > b) {a = b; return true; } return false; } template bool chmax(T &a, const T& b) { if (a < b) {a = b; return true; } return false; } template void print(const T& t, const Ts&... ts) { cout << t; ((cout << ' ' << ts), ...); cout << '\n'; } constexpr static struct PositiveInfinity { template constexpr operator T() const { return numeric_limits::max() / 2; } constexpr auto operator-() const; } inf; // NOLINT constexpr static struct NegativeInfinity { template constexpr operator T() const { return numeric_limits::lowest() / 2; } constexpr auto operator-() const; } NegativeInfinityVal; constexpr auto PositiveInfinity::operator-() const { return NegativeInfinityVal; } constexpr auto NegativeInfinity::operator-() const { return inf; } // clang-format on ll modpow(ll a, ll n, ll mod) { if (n == 0) return 1; if (n % 2 == 0) { ll res = modpow(a, n / 2, mod); return (res * res) % mod; } return (a * modpow(a, n - 1, mod)) % mod; } int main() { cin.tie(nullptr); ios::sync_with_stdio(false); int n, p, q; cin >> n >> p >> q; vector a(n); rep(i, 0, n) cin >> a[i]; vector cnt7(n, 0), cnt9(n, 0), cnt10(n, 0); rep(i, 0, n) { cnt7[i] = modpow(7, a[i], p); cnt9[i] = modpow(9, a[i], p); cnt10[i] = modpow(10, a[i], p); } auto cnt5 = make_v(n + 1, p + 1, 0); for (int i = n - 1; i >= 0; i--) { ll v = modpow(5, a[i], p); cnt5[i] = cnt5[i + 1]; cnt5[i][v]++; } ll ans = 0; rep(i, 0, n) { rep(j, i + 1, n) { rep(k, j + 1, n) { ll tmp = (cnt10[i] + cnt9[j] + cnt7[k]) % p; ll r = (q - tmp + p) % p; ans += cnt5[k + 1][r]; } } } print(ans); return 0; }