// 有意思但是 trivial 的一道数论题。
// 复杂度分析稍有些麻烦，咕咕咕了。
// 答案不会爆 ull，所以就直接自然溢出，啥事没有。

#include <algorithm>
#include <map>
#include <stdio.h>
#include <vector>
typedef long long llt;
typedef unsigned uint;typedef unsigned long long ullt;
typedef bool bol;typedef char chr;typedef void voi;
typedef double dbl;
template<typename T>bol _max(T&a,T b){return(a<b)?a=b,true:false;}
template<typename T>bol _min(T&a,T b){return(b<a)?a=b,true:false;}
template<typename T>T lowbit(T n){return n&-n;}
template<typename T>T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<typename T>T lcm(T a,T b){return(a!=0||b!=0)?a/gcd(a,b)*b:(T)0;}
template<typename T>T exgcd(T a,T b,T&x,T&y){if(b!=0){T ans=exgcd(b,a%b,y,x);y-=a/b*x;return ans;}else return y=0,x=1,a;}
template<typename T>T power(T base,T index,T mod)
{
    T ans=1%mod;
    while(index)
    {
        if(index&1)ans=ans*base%mod;
        base=base*base%mod,index>>=1;
    }
    return ans;
}
const uint Lim=1e6;
std::vector<uint>Fac[2005];
ullt Mu[Lim+5];bol Gone[Lim+5];
std::vector<ullt>_G[2005];
std::map<uint,ullt>Ans[2005];
ullt H(uint n)
{
    if(n<=Lim)return _G[1][n];
    if(Ans[1].count(n))return Ans[1][n];
    ullt ans=1;
    for(uint l=2,r;l<=n;l=r+1)r=n/(n/l),ans-=(r-l+1)*H(n/l);
    return Ans[1][n]=ans;
}
ullt G(uint n,uint d)
{
    if(n<=Lim/d)return _G[d][n];
    if(d==1)return H(n);
    if(Ans[d].count(n))return Ans[d][n];
    ullt ans=0;
    for(auto k:Fac[d])ans+=Mu[k]*G(n/k,k);
    ans*=Mu[d];
    return Ans[d][n]=ans;
}
std::map<uint,ullt>_F[2005];
ullt F(uint n,uint d)
{
    if(_F[d].count(n))return _F[d][n];
    ullt ans=0;
    for(auto k:Fac[d])ans+=Mu[k]*(n/k);
    return _F[d][n]=ans;
}
int main()
{
#ifdef MYEE
    freopen("QAQ.in","r",stdin);
    // freopen("QAQ.out","w",stdout);
#endif
    Mu[1]=1;
    for(ullt i=2,tp=0;i<=Lim;i++){
        static ullt Prime[1000005];if(!Gone[i]){Mu[i]=-1llu,Prime[tp++]=i;for(ullt j=i*i;j<=Lim;j*=i)Gone[j]=true;}
        for(auto p:Prime)if(i%p&&i*p<=Lim){Mu[i*p]=-Mu[i];for(ullt j=p;i*j<=Lim;j*=p)Gone[i*j]=true;}else break;
    }
    for(uint i=1;i<=2000;i++)
    {
        for(uint j=i;j<=2000;j+=i)Fac[j].push_back(i);
        _G[i]={0};for(uint j=i;j<=Lim;j+=i)_G[i].push_back(_G[i].back()+Mu[j]);
    }
    uint n,m,k;scanf("%u%u%u",&n,&m,&k);
    ullt ans=0;
    for(auto b:Fac[k])if(Mu[b])for(auto h:Fac[b])for(uint l=1,r;(ullt)l*h<=n&&(ullt)l*b<=m;l=r+1)
    {
        r=std::min(n/h/(n/h/l),m/b/(m/b/l));
        ans+=(G(r,h)-G(l-1,h))*(n/h/l)*F(m/b/l,h)*Mu[b];
    }
    printf("%llu\n",ans);
    return 0;
}