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using namespace std;
#define ll long long
const int maxn = 35;
const int mod = 8191;
int n, k;
ll x1, y1;
int p;
struct
{
int ma[4][4];
martix friend operator *(const martix a, const martix b){
martix ret;
memset(ret.ma,0,sizeof(ret.ma));
for(int i=0;i<p;i++){
for(int j=0;j<p;j++){
for(int k=0;k<p;k++)
ret.ma[i][j]=(ret.ma[i][j] + a.ma[i][k]*b.ma[k][j]%mod)%mod;
}
}
return ret;
}
}ans;
martix multipow(martix x, int k)
{
memset(ans.ma, 0, sizeof(ans.ma));
for(int i=0;i<p;i++)
ans.ma[i][i]=1;
while(k)
{
if(k&1) ans=ans*x;
k>>=1;
x=x*x;
}
return ans;
}
void seek(ll &x, ll &y, ll d)
{
y=1;
while(1){
x=1LL*sqrt(d*y*y+1);
if(x*x-d*y*y==1) break;
y++;
}
}
bool check(int x)
{
int y=(int)sqrt(x);
if(y*y==x) return true;
return false;
}
int main()
{
p=2;
while(scanf("%d%d", &n, &k)!=EOF)
{
if(check(n)) {
printf("No answers can meet such conditionsn");
continue;
}
seek(x1, y1, n);
martix now;
now.ma[0][0]=x1;
now.ma[0][1]=(n)*(y1);
now.ma[1][0]=y1;
now.ma[1][1]=x1;
martix ans=multipow(now, k-1);
int sum=0;
sum = (sum + ans.ma[0][0]*x1%mod + ans.ma[0][1]*y1%mod) % mod;
printf("%dn", sum);
}
}
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