#include <iostream>
#include <vector>
using namespace std;

inline long long mod(long long a, long long m) {
    return (a % m + m) % m;
}

// 拡張 Euclid の互除法
long long extGCD(long long a, long long b, long long &p, long long &q) {
    if (b == 0) { p = 1; q = 0; return a; }
    long long d = extGCD(b, a%b, q, p);
    q -= a/b * p;
    return d;
}

// 逆元
long long modinv(long long a, long long m) {
    long long x, y;
    extGCD(a, m, x, y);
    return mod(x, m); // 気持ち的には x % m だが、x が負かもしれないので
}

// 中国剰余定理
pair<long long, long long> ChineseRem(long long b1, long long m1, long long b2, long long m2) {
    long long p, q;
    long long d = extGCD(m1, m2, p, q); // p is inv of m1/d (mod. m2/d)
    if ((b2 - b1) % d != 0) return make_pair(0, -1);
    long long m = m1 * (m2/d);
    long long tmp = (b2 - b1) / d * p % (m2/d);
    long long r = mod(b1 + m1 * tmp, m);
    return make_pair(r, m);
}

int n, m;
const int MOD[2] = {168647939, 592951213};
long long b[2];
vector<int> x;
int dp[10000010];

int main() {
    cin >> n >> m; --n;
    x.resize(m); for (int i = 0; i < m; ++i) cin >> x[i];
    for (int iter = 0; iter < 2; ++iter) {
        int p = MOD[iter];
        dp[0] = 1;
        for (int i = 1; i <= (n+1)/2; ++i) {
            long long tmp = 0;
            for (auto xv : x) if (i >= xv) (tmp += dp[i - xv]) %= p;
            dp[i] = (int)tmp;
        }
        b[iter] = 0;
        for (int i = 0; i <= (n+1)/2; ++i) for (auto xv : x) {
            if (i + xv > (n+1)/2 && i + xv <= n) {
                (b[iter] += (long long)(dp[i]) * dp[n - (i + xv)]) %= p;
            }
        }
    }
    pair<long long, long long> res = ChineseRem(b[0], MOD[0], b[1], MOD[1]);
    cout << res.first << endl;
}