Dr. Ceizenp'ok from planet i1c5l became famous across the whole Universe thanks to his recent discovery — the Ceizenpok’s formula. This formula has only three arguments: n, k and m, and its value is a number of k-combinations of a set of n modulo m.

While the whole Universe is trying to guess what the formula is useful for, we need to automate its calculation.

Single line contains three integers n, k, m, separated with spaces (1 ≤ n ≤ 1018, 0 ≤ k ≤ n, 2 ≤ m ≤ 1 000 000).

Write the formula value for given arguments n, k, m.

/*
2015 ICL, Finals, Div. 1 Ceizenpok’s formula（组合数取模，扩展lucas定理）求C(n,k)%m
如果m是素数的话直接就能套lucas模板.
对于m为合数,我们可以把它分解成素数在进行处理 m = p1*p2*p3..pk   (pk = prime[i]^t)
然后利用扩展lucas定理可以求出  C(n,k) % pi的值，最后利用中国剩余定理涨姿势：http://www.cnblogs.com/jianglangcaijin/p/3446839.html
题目链接：http://codeforces.com/gym/100633/problem/J
hhh-2016-04-16 13:07:05
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <functional>
typedef long long ll;
#define lson (i<<1)
#define rson ((i<<1)|1)
using namespace std;const int maxn = 1e6+10;
ll fac[maxn];
int w[maxn],num[maxn],tw[maxn];
int tot;
void get_factor(ll m)
{ll mm = m;tot = 0;for(ll i = 2; i*i <= m; i++){if(mm % i == 0){num[tot] = 0;w[tot] = i;tw[tot] = 1;while(mm % i == 0){num[tot]++;mm /= i;tw[tot] *= i;}tot++;}}if(mm > 1){num[tot] = 1;w[tot] = mm;tw[tot] = mm;tot ++;}
}ll ex_gcd(ll a,ll b,ll &x,ll &y)
{if(a == 0 && b == 0)return -1;if(b == 0){x = 1,y = 0;return a;}ll d = ex_gcd(b,a%b,y,x);y -= a/b*x;return d;
}ll pow_mod(ll a,ll b,ll mod)
{ll ret = 1;while(b){if(b&1) ret = ret*a%mod;a = a*a%mod;b >>= 1;}return ret;
}ll revers(ll a,ll b)
{ll x,y;ll d = ex_gcd(a,b,x,y);if(d == 1) return (x%b+b)%b;else return 0;
}ll c1(ll n,ll p,ll pk)
{if(n==0)return 1;ll ans=1;for(ll i = 2; i <= pk; i++)if(i % p)ans = ans*i%pk;ans=pow_mod(ans,n/pk,pk);for(ll k=n%pk,i=2; i<=k; i++)if(i%p)ans=ans*i%pk;return ans*c1(n/p,p,pk)%pk;
}ll cal(ll n,ll m,int cur,ll mod)
{ll pi = w[cur],pk = tw[cur];ll k = 0,ans;ll a,b,c;a=c1(n,pi,pk),b=c1(m,pi,pk),c=c1(n-m,pi,pk);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*revers(b,pk)%pk*revers(c,pk)%pk*pow_mod(pi,k,pk)%pk;return ans*(mod/pk)%mod*revers(mod/pk,pk)%mod;
}ll lucas(ll n,ll m,ll mod)
{ll ans = 0;for(int i = 0; i < tot; i++){ans = (ans+cal(n,m,i,mod))%mod;}return ans;
}ll n,k;
ll m;
int main()
{int T;while(scanf("%I64d%I64d%I64d",&n,&k,&m) != EOF){get_factor(m);printf("%I64d\n",lucas(n,k,m));}return 0;
}