#include #include #define rep(i,n) for(int i=0;i vi; typedef vector vl; typedef vector> vvi; typedef vector> vvl; typedef long double ld; typedef pair P; ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;} template ostream& operator<<(ostream& os, const static_modint& a) {os << a.val(); return os;} template ostream& operator<<(ostream& os, const dynamic_modint& a) {os << a.val(); return os;} template istream& operator>>(istream& is, vector& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;} template ostream& operator<<(ostream& os, const pair& p){os << p.first << ' ' << p.second; return os;} template ostream& operator<<(ostream& os, const vector& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;} template ostream& operator<<(ostream& os, const vector>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;} template ostream& operator<<(ostream& os, const set& se){for(T x : se) os << x << " "; os << "\n"; return os;} template ostream& operator<<(ostream& os, const unordered_set& se){for(T x : se) os << x << " "; os << "\n"; return os;} template ostream& operator<<(ostream& os, const atcoder::segtree& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template ostream& operator<<(ostream& os, const atcoder::lazy_segtree& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template void chmin(T& a, T b){a = min(a, b);} template void chmax(T& a, T b){a = max(a, b);} using mint = modint998244353; int main(){ int x, k; cin >> x >> k; vector> dp(2 * k + 1, vector(2 * k + 1)); dp[0][0] = 1; mint p = mint(x) / 100; mint q = mint(100 - x) / 100; rep(i, 2 * k){ vector> dp_old(2 * k + 1, vector(2 * k + 1)); swap(dp, dp_old); rep(from, 2 * k){ rep(from_max, 2 * k){ dp[from + 1][max(from_max, from + 1)] += dp_old[from][from_max] * p; if(from > 0) dp[from - 1][from_max] += dp_old[from][from_max] * q; } } } mint ans = 0; rep(from_max, 2 * k) ans += dp[0][from_max] * from_max; cout << ans << "\n"; return 0; }