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//法1：费马小定理求逆元，p需为质数 :
typedef long long ll;
const int MOD = 9973;

ll PowerMod(ll a, ll b, ll p) { //(a^b)%p
	ll ans = 1;
	a = a%p;
	while (b>0) {
		if (b & 1) ans = (ans*a) % p;
		b >>= 1;
		a = (a*a) % p;
	}
	return ans;
}

int main()
{
	ll n, b;
	scanf("%d%d", &n, &b);  //求(n/b)%MOD 
	printf("%lld\n", (n*PowerMod(b, MOD - 2, MOD)) % MOD);
	return 0;
}