#include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; typedef long long ll; const int inf=1<<30; const ll INF=1LL<<62; typedef pair P; typedef pair PP; const ll MOD=998244353; const int dy[]={0,1,0,-1}; const int dx[]={1,0,-1,0}; const int mod = 998244353; struct mint { long long x; // typedef long long long long; mint(long long x=0):x((x%mod+mod)%mod){} mint operator-() const { return mint(-x);} mint& operator+=(const mint a) { if ((x += a.x) >= mod) x -= mod; return *this; } mint& operator-=(const mint a) { if ((x += mod-a.x) >= mod) x -= mod; return *this; } mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;} mint operator+(const mint a) const { return mint(*this) += a;}//後ろのconstはメンバxを変更しないことを示す mint operator-(const mint a) const { return mint(*this) -= a;} mint operator*(const mint a) const { return mint(*this) *= a;} bool operator==(const mint a) const {return a.x==x;} mint pow(unsigned long long int t) const { assert(t>=0); if (!t) return 1; //aがtで割り切れない場合には t%=(mod-1)をして良い mint a = pow(t>>1); a *= a; if (t&1) a *= *this; return a; } // for prime mod mint inv() const { return pow(mod-2);} mint& operator/=(const mint a) { return *this *= a.inv();} mint operator/(const mint a) const { return mint(*this) /= a;} }; std::istream& operator>>(std::istream& is, mint& a) { return is >> a.x;} std::ostream& operator<<(std::ostream& os, const mint& a) { return os << a.x;} struct combination { std::vector fact, ifact; combination(int n):fact(n+1),ifact(n+1) { assert(n < mod); fact[0] = 1; for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i; ifact[n] = fact[n].inv(); for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i; } mint operator()(int n, int k) { if (k < 0 || k > n) return 0; return fact[n]*ifact[k]*ifact[n-k]; } }; int main(){ int x,K; cin>>x>>K; mint p=(mint)x/100; mint q=(mint)1-p; int len=2*K; mint ans=0; for(int S=0;S<(1<>i)&1){ //( c++; max_depth=max(max_depth,c); } else{ //) c--; if(c<0) ok=false; } } if(ok){ ans+=(mint)max_depth*p.pow(K)*q.pow(K); } } cout<