题目要求$g^{{\sum }_{d|n}{C}_n^d}mod999911659$
根据拓展欧拉定理
$g^{{\sum }_{d|n}{C}_n^{d}mod999911658}mod999911659$
把999911658分解质因数，得到2，3，4679， 35617
分别做lucas，然后用中国剩余定理和并

$code$

#include<iostream>
#include<cstdio>
#define ll long long

using namespace std;

ll P=999911658;
ll g, n, p[5]={0, 2, 3, 4679, 35617}, q[5];

ll qpow(ll x, ll t, ll mod)
{
	ll sum=1ll;
	while(t)
	{
		if(t&1)
			sum=sum*x%mod;
		x=x*x%mod;
		t>>=1;
	}
	return sum%mod;
}

ll inv(ll x, ll mod)
{
	return qpow(x, mod-2, mod);
}

ll C(ll n, ll m, ll mod)
{
	if(n<m)
		return 0;
	if(m==0)
		return 1;
	ll sum=1;
	for(ll i=n-m+1; i<=n; i++)
		sum=sum*i%mod;
	ll chu=1;
	for(ll i=1; i<=m; i++)
		chu=chu*i%mod;
	return sum*inv(chu, mod)%mod;
}

ll lucas(ll n, ll m, ll mod)
{
	if(n<m)
		return 0;
	if(m==0)
		return 1;
	return C(n%mod, m%mod, mod)*lucas(n/mod,m/mod, mod);
}

void a()
{
	for(ll i=1; i*i<=n; i++)
	{
		if(n%i==0)
		{
			for(ll j=1; j<=4; j++)
				q[j]=(q[j]+lucas(n, i, p[j]))%p[j];
			if(i*i!=n)
			{
				for(ll j=1; j<=4; j++)
					q[j]=(q[j]+lucas(n, n/i, p[j]))%p[j];
			}
		}
	}
}

ll b()
{
	ll ans=0, mo1=P;
	for(ll i=1; i<=4; i++)
	{
		ll m=mo1/p[i];
		ll t=inv(m, p[i]);
		ans=(ans+m*t*q[i])%mo1;
	}
	return ans;
}

int main()
{
	scanf("%lld%lld", &n, &g);
	g%=P+1;
	if(g==0)
	{
		printf("0");
		return 0;
	}
	a();
	ll k=(b()+P)%P;
	printf("%lld", qpow(g, k, P+1));
	return 0;
}