ll pow(ll a, ll b, ll m)

 {

     ll ans = ;

     a %= m;

     while(b)

     {

         if(b & )ans = (ans % m) * (a % m) % m;

         b /= ;

         a = (a % m) * (a % m) % m;

     }

     ans %= m;

     return ans;

 }

 ll inv(ll x, ll p)//x关于p的逆元，p为素数

 {

     return pow(x, p - , p);

 }

 ll C(ll n, ll m, ll p)//组合数C(n, m) % p

 {

     if(m > n)return ;

     ll up = , down = ;//分子分母;

     for(int i = n - m + ; i <= n; i++)up = up * i % p;

     for(int i = ; i <= m; i++)down = down * i % p;

     return up * inv(down, p) % p;

 }

 ll Lucas(ll n, ll m, ll p)

 {

     if(m == )return ;

     return C(n % p, m % p, p) * Lucas(n / p, m / p, p) % p;

 }

 const int maxn = 1e5 + ;

 ll fac[maxn];//阶乘打表

 void init(ll p)//此处的p应该小于1e5，这样Lucas定理才适用

 {

     fac[] = ;

     for(int i = ; i <= p; i++)

         fac[i] = fac[i - ] * i % p;

 }

 ll pow(ll a, ll b, ll m)

 {

     ll ans = ;

     a %= m;

     while(b)

     {

         if(b & )ans = (ans % m) * (a % m) % m;

         b /= ;

         a = (a % m) * (a % m) % m;

     }

     ans %= m;

     return ans;

 }

 ll inv(ll x, ll p)//x关于p的逆元，p为素数

 {

     return pow(x, p - , p);

 }

 ll C(ll n, ll m, ll p)//组合数C(n, m) % p

 {

     if(m > n)return ;

     return fac[n] * inv(fac[m] * fac[n - m], p) % p;

 }

 ll Lucas(ll n, ll m, ll p)

 {

     if(m == )return ;

     return C(n % p, m % p, p) * Lucas(n / p, m / p, p) % p;

 }

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 const int maxn = 1e6 + ;

 const int mod = 1e9 + ;

 ll pow(ll a, ll b, ll m)

 {

     ll ans = ;

     a %= m;

     while(b)

     {

         if(b & )ans = (ans % m) * (a % m) % m;

         b /= ;

         a = (a % m) * (a % m) % m;

     }

     ans %= m;

     return ans;

 }

 ll extgcd(ll a, ll b, ll& x, ll& y)

 //求解ax+by=gcd(a, b)

 //返回值为gcd(a, b)

 {

     ll d = a;

     if(b)

     {

         d = extgcd(b, a % b, y, x);

         y -= (a / b) * x;

     }

     else x = , y = ;

     return d;

 }

 ll mod_inverse(ll a, ll m)

 //求解a关于模上m的逆元

 //返回-1表示逆元不存在

 {

     ll x, y;

     ll d = extgcd(a, m, x, y);

     return d ==  ? (m + x % m) % m : -;

 }

 ll Mul(ll n, ll pi, ll pk)//计算n! mod pk的部分值  pk为pi的ki次方

 //算出的答案不包括pi的幂的那一部分

 {

     if(!n)return ;

     ll ans = ;

     if(n / pk)

     {

         for(ll i = ; i <= pk; i++) //求出循环节乘积

             if(i % pi)ans = ans * i % pk;

         ans = pow(ans, n / pk, pk); //循环节次数为n / pk

     }

     for(ll i = ; i <= n % pk; i++)

         if(i % pi)ans = ans * i % pk;

     return ans * Mul(n / pi, pi, pk) % pk;//递归求解

 }

 ll C(ll n, ll m, ll p, ll pi, ll pk)//计算组合数C(n, m) mod pk的值 pk为pi的ki次方

 {

     if(m > n)return ;

     ll a = Mul(n, pi, pk), b = Mul(m, pi, pk), c = Mul(n - m, pi, pk);

     ll k = , ans;//k为pi的幂值

     for(ll i = n; i; i /= pi)k += i / pi;

     for(ll i = m; i; i /= pi)k -= i / pi;

     for(ll i = n - m; i; i /= pi)k -= i / pi;

     ans = a * mod_inverse(b, pk) % pk * mod_inverse(c, pk) % pk * pow(pi, k, pk) % pk;//ans就是n! mod pk的值

     ans = ans * (p / pk) % p * mod_inverse(p / pk, pk) % p;//此时用剩余定理合并解

     return ans;

 }

 ll Lucas(ll n, ll m, ll p)

 {

     ll x = p;

     ll ans = ;

     for(ll i = ; i <= p; i++)

     {

         if(x % i == )

         {

             ll pk = ;

             while(x % i == )pk *= i, x /= i;

             ans = (ans + C(n, m, p, i, pk)) % p;

         }

     }

     return ans;

 }

 int main()

 {

     ll n, m, p;

     while(cin >> n >> m >> p)

     {

         cout<<Lucas(n, m, p)<<endl;

     }

     return ;

 }