#include <bits/stdc++.h>

using namespace std;
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)
#define rrep(i,n) for(int i=n-1;i>=0;--i)
#define debug(output) if(debugFlag)cout<<#output<<"= "<<output<<"\n";
using lint = long long;
typedef pair<int,int> P;
const bool debugFlag=true;
const lint linf=1.1e18;const int inf=1.01e9;
constexpr int MOD=1000000007;
template<class T>bool chmax(T &a, const T &b) { if(a < b){ a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if(a > b){ a = b; return 1; } return 0; }

#include<atcoder/all>
using namespace atcoder;
using mint = modint998244353;

template<class T>
struct Binom{
    long long n;
    bool flag=false;
    vector<T> kaijo,inv_kaijo;
    Binom(int N){
        n=N;
        kaijo.push_back(1);
        for(long long i=1;i<=n;i++){
            kaijo.push_back((kaijo[i-1]*i));
        }
    }
    void invList(){
      inv_kaijo.push_back(1);
      flag=true;
      for(int i=1;i<=n;++i){
        inv_kaijo.push_back(1/kaijo[i]);
      }
    }
    T binom(long long N,long long K){
        if(N<0||K<0)return 0;
        if(N<K)return 0;
        if(flag)return kaijo[N]*inv_kaijo[K]*inv_kaijo[N-K];
        T res=kaijo[N]/kaijo[K];//
        res/=kaijo[N-K];
        return res;
    }
    T binom2(mint N,long long K){
        // O(K)で2項係数を返す
        T res=1/kaijo[K];
        for(long long i=0;i<K;++i){
            res*=(N-i);
        }
        return res;
    }
};

signed main(){
    int n;cin>>n;
    vector<int> a;
    rep(i,9)a.push_back(i);
    a.push_back(8);
    for(int i=9;i<98;++i)a.push_back(i);
    a.push_back(97);
    for(int i=98;i<=200;++i)a.push_back(i);
    vector<vector<mint>> dp(n+1,vector<mint>(n,0));
    dp[0][0]=1;
    Binom<mint> binom(n+1000);
    binom.invList();
    for(auto& x:a){
        auto ndp=dp;
        rep(i,n){
            rep1(j,n){
                if(i+j>n)break;
                if(x*j>=n)break;
                rep(k,n){
                    if(k+x*j>=n)break;
                    ndp[i+j][k+x*j]+=dp[i][k]*binom.binom(n-i,j);
                }
            }
        }
        swap(dp,ndp);
    }
    cout<<dp[n][n-1].val()<<"\n";
    return 0;
}