Problem

There are coins with front and back sides and dice with rolls from $ 1 $ to $ N $. Gacho decided to play the following games using these.

The game starts with a score of $ 0 $ and proceeds as follows.


1. Roll the dice $ 1 $ and add the number of rolls to the score.
2. If the current score is $ K $ or more, the game is cleared and the game ends.
3. If the current score is less than $ K $, toss a coin and return to 1. if the front appears, and if the back appears, the game is over and the game ends.



When a coin is thrown, it has a probability of $ A \% $ and a probability of $ (100-A) \% $. Also, when the dice are rolled, each eye appears with equal probability.
At this time, find the probability that Gacho can clear the game in one game.
When the probability to be calculated is expressed as $ \ frac {P} {Q} $ using relatively prime integers $ P and Q $, it becomes $ R \ times Q \ equiv P \ bmod 998244353 $ $ 0 $ or more and $ 998244352 $ or less. Output the integer $ R $ of. Under the constraints of this problem, such $ R $ always exists uniquely.

Constraints

The input satisfies the following conditions.

* $ 1 \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq K \ leq 10 ^ 5 $
* $ 1 \ leq A \ leq 99 $
* All inputs are integers

Input

The input is given in the following format.


$ N $ $ K $ $ A $


$ N, K, A $ are given on one line separated by blanks.

Output

When the probability of clearing the game is expressed as $ \ frac {P} {Q} $ using relatively prime integers $ P and Q $, it becomes $ R \ times Q \ equiv P \ bmod 998244353 $. Output an integer $ R $ that is greater than or equal to $ 0 $ and less than or equal to $ 998244352 $.

Examples

Input

1 1 50


Output

1


Input

2 2 10


Output

648858830


Input

6 10 99


Output

650893870
#include <bits/stdc++.h>
#define whlie while
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define rep(i,N) for(int i = 0; i < (N); i++)
#define repr(i,N) for(int i = (N) - 1; i >= 0; i--)
#define rep1(i,N) for(int i = 1; i <= (N) ; i++)
#define repr1(i,N) for(int i = (N) ; i > 0 ; i--)
#define each(x,v) for(auto& x : v)
#define all(v) (v).begin(),(v).end()
#define sz(v) ((int)(v).size())
#define vrep(v,it) for(auto it = v.begin(); it != v.end(); it++)
#define vrepr(v,it) for(auto it = v.rbegin(); it != v.rend(); it++)
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) ll __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
using namespace std; void solve();
using ll = long long; using vl = vector<ll>;
using vi = vector<int>; using vvi = vector< vector<int> >;
constexpr int inf = 1001001001;
constexpr ll infLL = (1LL << 61) - 1;
struct IoSetupNya {IoSetupNya() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); cerr << fixed << setprecision(7);} } iosetupnya;
template<typename T, typename U> inline bool amin(T &x, U y) { return (y < x) ? (x = y, true) : false; }
template<typename T, typename U> inline bool amax(T &x, U y) { return (x < y) ? (x = y, true) : false; }
template<typename T, typename U> ostream& operator <<(ostream& os, const pair<T, U> &p) { os << p.first << " " << p.second; return os; }
template<typename T, typename U> istream& operator >>(istream& is, pair<T, U> &p) { is >> p.first >> p.second; return is; }
template<typename T> ostream& operator <<(ostream& os, const vector<T> &v) { int s = (int)v.size(); rep(i,s) os << (i ? " " : "") << v[i]; return os; }
template<typename T> istream& operator >>(istream& is, vector<T> &v) { for(auto &x : v) is >> x; return is; }
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << "\n";} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
template<typename T>void die(T x){out(x); exit(0);}
#ifdef NyaanDebug
  #include "NyaanDebug.h"
  #define trc(...) do { cerr << #__VA_ARGS__ << " = "; dbg_out(__VA_ARGS__);} while(0)
  #define trca(v,...) do { cerr << #v << " = "; array_out(v , __VA_ARGS__ );} while(0)
#else
  #define trc(...)
  #define trca(...)
  int main(){solve();}
#endif
 
constexpr int MOD = /** 1000000007; //*/ 998244353;
/////////////////

// BIT

template< typename T >
struct BIT {
  int N; int max_2beki;

  vector< T > data;
  // 初期化 1-indexedでデータを管理する 0で初期化
  BIT(int size){
      N = ++size;
      data.assign(N, 0);
      max_2beki = 1;
      while(max_2beki * 2 <= N) max_2beki *= 2;
  }

  // [0,k](閉区間)の総和 閉区間に注意！
  T sum(int k) {
    if(k < 0) return 0; // k<0のとき0を返す
    T ret = 0;
    for(++k; k > 0; k -= k & -k) ret += data[k];
    return (ret);
  }

  // [l,r](閉区間)の総和
  inline T sum(int l,int r){
    return sum(r) - sum(l-1);
  }

  // 一点取得 更新はできないことに注意
  inline T operator[](int k){
    return sum(k) - sum(k-1);
  }

  // data[k] += x;
  void add(int k, T x) {
    for(++k; k < N; k += k & -k) data[k] += x;
  }

  // imos法 [l,r]にxを加算
  void imos(int l,int r,T x){
    add(l , x); add(r + 1 , -x);
  }

  // lower_bound sum(i)がval以上となる最小のi
  int lower_bound(T w){
    if(w <= 0) return 0;
    int x = 0;
    for(int k = max_2beki; k > 0; k /= 2){
      if(x+k <= N - 1 && data[x + k] < w){
        w -= data[x + k];
        x += k;
      }
    }
    return x;
  }

  // upper_bound sum(i)がvalより大きくなる最小のi
  int upper_bound(T w){
    if(w < 0) return 0;
    int x = 0;
    for(int k = max_2beki; k > 0; k /= 2){
      if(x+k <= N - 1 && data[x + k] <= w){
        w -= data[x + k];
        x += k;
      }
    }
    return x;
  }

};

template< int mod >
struct ModInt {
  int x;

  ModInt() : x(0) {}

  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

  ModInt &operator+=(const ModInt &p) {
    if((x += p.x) >= mod) x -= mod;
    return *this;
  }

  ModInt &operator-=(const ModInt &p) {
    if((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }

  ModInt &operator*=(const ModInt &p) {
    x = (int) (1LL * x * p.x % mod);
    return *this;
  }

  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }

  ModInt operator-() const { return ModInt(-x); }

  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }

  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }

  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }

  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }

  bool operator==(const ModInt &p) const { return x == p.x; }

  bool operator!=(const ModInt &p) const { return x != p.x; }

  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while(b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }

  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while(n > 0) {
      if(n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }

  friend ostream &operator<<(ostream &os, const ModInt &p) {
    return os << p.x;
  }

  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt< mod >(t);
    return (is);
  }

  static int get_mod() { return mod; }
};

using modint = ModInt< MOD >;

void solve(){
  int N,K,A;
  cin >> N >> K >> A;
  BIT<modint> bit(N + K + 10);
  bit.add(0 , 1);
  rep1(i , K + N - 1){
    trc(bit.sum( max(0,i-N) , min(K-1,i-1) ));
    bit.add(i , bit.sum( max(0,i-N) , min(K-1,i-1) ) * A / (N * 100));
    if(i >= K)
      bit.add(i , bit.sum( max(0,i-N) , min(K-1,i-1) ) * (100-A)/(N*100));
  }
  out(bit.sum(K , K + N - 1));
}
