#include <bits/stdc++.h>
using ll = long long;
constexpr ll MOD1 = 168647939;
constexpr ll MOD2 = 592951213;
template <typename T>
constexpr std::pair<T, T> extgcd(const T a, const T b)
{
    if (b == 0) { return std::pair<T, T>{1, 0}; }
    const auto p = extgcd(b, a % b);
    return {p.second, p.first - p.second * (a / b)};
}
template <typename T>
std::pair<T, T> ChineseRemainderTheorem(const std::pair<T, T>& a1, const std::pair<T, T>& a2)  // (mod, value)
{
    const T p1 = a1.first, m1 = a1.second, p2 = a2.first, m2 = a2.second, m = p1 * p2;
    if (m1 == m2) { return {p1 * p2, m1}; }
    auto p = extgcd(p1, p2);
    return {m, (((p1 * p.first * (m2 - m1) + m1) % m) + m) % m};
}
int main()
{
    int N, M;
    std::cin >> N >> M;
    std::vector<int> x(M);
    for (int i = 0; i < M; i++) { std::cin >> x[i]; }
    const int H = (N + 1) / 2;
    auto solve = [&](const ll mod) {
        std::vector<int> dp(H, 0);
        dp[0] = 1;
        for (int i = 1; i < H; i++) {
            for (int j = 0; j < M; j++) {
                if (i >= x[j]) { (dp[i] += dp[i - x[j]]) %= mod; }
            }
        }
        ll ans = 0;
        for (int j = 0; j < M; j++) {
            for (int i = std::max(0, x[j] - H); i < std::min(N - H, x[j]); i++) { (ans += (ll)dp[H - x[j] + i] * dp[N - H - i - 1] % mod) %= mod; }
        }
        return ans;
    };
    const auto ans = ChineseRemainderTheorem<__int128_t>({MOD1, solve(MOD1)}, {MOD2, solve(MOD2)});
    std::cout << (ll)ans.second << std::endl;
    return 0;
}