#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define FORR2(x,y,arr) for(auto& [x,y]:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
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;}
//-------------------------------------------------------

int N,K;
int A[101];
ll from[10101];

ll dp[5100][101][101];
const ll mo=998244353;

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N>>K;
	map<int,int> M;
	FOR(i,N) {
		cin>>x;
		M[-x]++;
	}
	int cur=0;
	from[0]=1;
	FORR2(e,n,M) {
		FOR(x,5051) {
			FOR(i,cur+1) FOR(j,n+1) dp[x][i][j]=0;
			dp[x][0][0]=from[x];
		}
		FOR(x,5051) {
			FOR(i,cur+1) FOR(j,n+1) if(dp[x][i][j]) {
				if(i<cur) (dp[x][i+1][j]+=dp[x][i][j])%=mo;
				if(j<n) (dp[x+i][i][j+1]+=dp[x][i][j])%=mo;
			}
		}
		FOR(x,5051) from[x]=dp[x][cur][n];
		cur+=n;
	}
	cout<<from[K]<<endl;
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}