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#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (b); i >= (a); i--)
#define SZ(x) ((int)x.size())
#define all(x) x.begin(), x.end()
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define st first
#define nd second
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
auto &operator<<(auto &o, pair<auto, auto> p) {
    return o << "(" << p.st << ", " << p.nd << ")";
}
auto operator<<(auto &o, auto x)->decltype(end(x), o) {
    o << "{"; int i=0; for(auto e: x) o << ", " + 2*!i++ << e;
    return o << "}";
}
#define deb(x...) cerr << "[" #x "]: ", [](auto...$) { ((cerr<<$<<"; "),...) << endl; }(x)

const int nax = 105;
ll h, w;
ll a[nax];
int n;

ll cost(ll r1, ll r2, int id){
    if(id == n){
        return (r1 * r2 / (a[id] * a[id]));
    }
    ll nxtr1 = (r1 / a[id + 1] * a[id + 1]);
    ll nxtr2 = (r2 / a[id + 1] * a[id + 1]);
    ll ans = (r1 * r2 - nxtr1 * nxtr2) / (a[id] * a[id]);
    return ans + cost(nxtr1, nxtr2, id + 1);
}

void solve(){
    cin >> h >> w;
    cin >> n;
    rep(i, 1, n) cin >> a[i];
    if(h % a[1] != 0 || w % a[1] != 0){
        cout << -1 << "\n";
        return;
    }
    cout << cost(h, w, 1) << "\n";
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int tt = 1;
    // cin >> tt;
    rep(te, 1, tt) solve();
    return 0;
}