#pragma GCC optimization ("O3") #include using namespace std; using ll = long long; using vec = vector; using mat = vector; using pll = pair; using dvec = vector; using dmat = vector; #define INF (1LL<<61) #define MOD 1000000007LL //#define MOD 998244353LL #define PR(x) cout << (x) << endl #define PS(x) cout << (x) << " " #define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i)) #define FORE(i,v) for(auto (i):v) #define ALL(x) (x).begin(), (x).end() #define SZ(x) ((ll)(x).size()) #define REV(x) reverse(ALL((x))) #define ASC(x) sort(ALL((x))) #define DESC(x) {ASC((x)); REV((x));} #define BIT(s,i) (((s)>>(i))&1) #define pb push_back #define fi first #define se second template inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;} template inline int chmax(T& a, T b) {if(a=MOD) x-=MOD; return *this;} mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;} mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;} mint operator+(const mint& a) const {mint b(*this); return b+=a;} mint operator-(const mint& a) const {mint b(*this); return b-=a;} mint operator*(const mint& a) const {mint b(*this); return b*=a;} mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;} mint inv() const {return pow(MOD-2);} mint& operator/=(const mint& a) {return *this*=a.inv();} mint operator/(const mint& a) const {mint b(*this); return b/=a;} }; istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;} ostream &operator<<(ostream& os, const mint& a) {return os<; using mmat = vector; vector factorize(ll n) { vector p; ll t = n; for(ll i=2; i*i<=t; ++i) { ll cnt = 0; if(t%i) continue; while(t%i == 0) { t /= i; ++cnt; } p.pb({i, cnt}); } if(t != 1) p.pb({t, 1}); return p; } ll legendre(ll n, ll p) { ll ret = 0, d = p; while(1) { ret += n/d; if(d > n/p) break; d *= p; } return ret; } int main() { ll N, K, M; cin >> N >> K >> M; ll ans = INF; vector P = factorize(M); FORE(p,P) { ll c = legendre(N, p.fi)-legendre(K, p.fi)-legendre(N-K, p.fi); chmin(ans, c/p.se); } PR(ans); return 0; } /* */