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
/* monkukui 競技プログラミング用のテンプレート  (ここから) */
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <functional>
#include <bitset>

using namespace std;
using lint = long long int;
using ll = long long int;
using lnt = long long int;
using graph = vector<vector<long long int>>;
using wgraph = vector<vector<pair<long long int, long long int>>>;
long long int INF = 1001001001001001LL;
long long int MOD = 998244353LL;
double PI = 3.1415926535897932;
long long int di[] = {-1, 0, 1, 0, -1, 1, 1, -1};
long long int dj[] = {0, 1, 0, -1, 1, 1, -1, -1};

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
inline void yes(){ cout << "yes" << endl; }
inline void Yes(){ cout << "Yes" << endl; }
inline void YES(){ cout << "YES" << endl; }
inline void no(){ cout << "no" << endl; }
inline void No(){ cout << "No" << endl; }
inline void NO(){ cout << "NO" << endl; }
inline void possible(){ cout << "possible" << endl; }
inline void Possible(){ cout << "Possible" << endl; }
inline void POSSIBLE(){ cout << "POSSIBLE" << endl; }
inline void impossible(){ cout << "impossible" << endl; }
inline void Impossible(){ cout << "Impossible" << endl; }
inline void IMPOSSIBLE(){ cout << "IMPOSSIBLE" << endl; }

#define rep(i,n) for(long long int i = 0; i < (n); i++)
#define rrep(i,n) for(long long int i = 1; i <= (n); i++)
#define drep(i,n) for(long long int i = (n)-1; i >= 0; i--)
#define srep(i,s,t) for(long long int i = s; i < t; i++)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define pb push_back

/* monkukui 競技プログラミング用のテンプレート (ここまで)*/

// quoted from beet-aizu
template <typename T, typename E, typename F, typename G>
struct SegmentTree{
    // using F = function<T(T, T)>
    // using G = function<T(T, E)>
    int n;
    F f;
    G g;
    T ti;
    vector<T> dat;
    SegmentTree(){};
    SegmentTree(F f,G g,T ti):f(f),g(g),ti(ti){}
    void init(int n_){    
        n=1;
        while(n<n_) n<<=1;
        dat.assign(n<<1,ti);
    }
    void build(const vector<T> &v){
        int n_=v.size();
        init(n_);
        for(int i=0;i<n_;i++) dat[n+i]=v[i];
        for(int i=n-1;i;i--)
            dat[i]=f(dat[(i<<1)|0],dat[(i<<1)|1]);
    }
    void update(int k,const E &x){
        k += n;
        dat[k] = g(dat[k], x);
        while(k>>=1)
            dat[k]=f(dat[(k<<1)|0],dat[(k<<1)|1]);    
    }
    T operator [](int k) const { return dat[k+n]; }
    T query(int a,int b) const {
        T vl=ti,vr=ti;
        for(int l=a+n,r=b+n;l<r;l>>=1,r>>=1) {
            if(l&1) vl=f(vl,dat[l++]);
            if(r&1) vr=f(dat[--r],vr);
        }
        return f(vl,vr);
    }

};


// 逆元を求める. a と m は互いに素であることが要請される.
lint modinv(lint a, lint m) {
    lint b = m, u = 1, v = 0;
    while(b){
        lint t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m; 
    if (u < 0) u += m;
    return u;
}

int main(){
  lint n, k, a; cin >> n >> k >> a;
  lint invn = modinv(n, MOD);
  lint inv100 = modinv(100, MOD);

  a = a * inv100;
  a %= MOD;

  using T = lint;  // type T
  using E = lint;  // type E
  auto f = [](T a, T b){ // return type T value
    return (a + b) % MOD;
  };
  auto g = [](T a, E b){ // return type T value
    return b % MOD;
  };
  T ti = 0;  // identify element
  SegmentTree<T, E, decltype(f), decltype(g)> sg(f, g, ti);  // don't change

  vector<lint> dat(n + k + 10, 0LL);
  dat[0] = 1LL;
  sg.build(dat);   // 初期配列を代入

  for(lint i = 1; i <= k + n + 2; i++) {
    lint l = max(0LL, i - n);
    lint r = min(k, i);
    lint sum = sg.query(l, r);
    sum = sum * invn % MOD;
    if(i < k) sum = sum * a % MOD;
    sg.update(i, sum);
  }

  cout << sg.query(k, k + n + 2) << endl;
}


