#include <bits/stdc++.h>
#include <atcoder/all>
#define OVERLOAD_REP(_1, _2, _3, name, ...) name
#define REP1(i,n) for (auto i = std::decay_t<decltype(n)>{}; (i) < (n); ++(i))
#define REP2(i,l,r) for (auto i = (l); (i) != (r); ++(i))
#define rep(...) OVERLOAD_REP(__VA_ARGS__, REP2, REP1)(__VA_ARGS__)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define pcnt __builtin_popcount
#define pcntl __builtin_popcountll
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using namespace atcoder;
using ll = long long;
using ull = unsigned long long;
using P = pair<int,int>;
using PP = pair<long long,int>;
using LP = pair<long long,long long>;
using mint1 = modint1000000007;
using mint9 = modint998244353;
const int INF = 1'100'000'000;
const long long LINF = 1'100'000'000'000'000'000LL;
const double DINF = 2e18;
const int di[] = {0,1,0,-1,-1,1,-1,1};
const int dj[] = {1,0,-1,0,1,1,-1,-1};
template<class T> inline bool chmin(T& a, T b){if(a>b){a=b;return true;}return false;}
template<class T> inline bool chmax(T& a, T b){if(a<b){a=b;return true;}return false;}
void YesNo (bool ok,const string& s="Yes",const string& t="No") {if(ok)cout<<s<<endl;else cout<<t<<endl;}


//-----def end-----//

long long Power (long long a , long long n , long long M) {
     if(n == 0) return 1;
     a%=M;
     if(n == 1) return a;
     if(n%2 == 0) {
         long long res = Power(a,n/2,M)%M;
         return (res*res)%M;
     } else {
         long long res = Power(a,n-1,M)%M;
         return (a*res)%M;
     }
 }

int main () {
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    long long n , p , q;
    cin >> n >> p >> q;
    vector<long long> a(n) , b(n) , c(n) , d(n) , e(n);
    rep(i , n) cin >> a[i];
    sort(all(a));
    long long res = 0;
    rep(i , n) b[i] = Power(10 , a[i] , p);
    rep(i , n) c[i] = Power(9 , a[i] , p);
    rep(i , n) d[i] = Power(7 , a[i] , p);
    rep(i , n) e[i] = Power(5 , a[i] , p);
    rep(i , n-3) rep(j , i+1 , n-2) rep(k , j+1 , n-1) rep(l , k+1 ,  n) {
        long long s = b[i]+c[j]+d[k]+e[l];
        s %= p;
        if(s == q) res++;
    }
    cout << res << endl;
    return 0;
}