Problem: Nezzar buys his favorite snack — n chocolate bars with lengths l_1,l_2,…,l_n. However, chocolate bars might be too long to store them properly! 

In order to solve this problem, Nezzar designs an interesting process to divide them into small pieces. Firstly, Nezzar puts all his chocolate bars into a black box. Then, he will perform the following operation repeatedly until the maximum length over all chocolate bars does not exceed k.

  * Nezzar picks a chocolate bar from the box with probability proportional to its length x. 
  * After step 1, Nezzar uniformly picks a real number r ∈ (0,x) and divides the chosen chocolate bar into two chocolate bars with lengths r and x-r. 
  * Lastly, he puts those two new chocolate bars into the black box. 



Nezzar now wonders, what is the expected number of operations he will perform to divide his chocolate bars into small pieces.

It can be shown that the answer can be represented as P/Q, where P and Q are coprime integers and Q not ≡ 0 (mod 998 244 353). Print the value of P⋅ Q^{-1} mod 998 244 353.

Input

The first line contains two integers n and k (1 ≤ n ≤ 50, 1 ≤ k ≤ 2000).

The second line contains n integers l_1, l_2, …, l_n (1 ≤ l_i, ∑_{i=1}^{n} l_i ≤ 2000).

Output

Print a single integer — the expected number of operations Nezzar will perform to divide his chocolate bars into small pieces modulo 998 244 353.

Examples

Input


1 1
2


Output


4


Input


1 1
1


Output


0


Input


1 5
1234


Output


15630811


Input


2 1
2 3


Output


476014684


Input


10 33
10 20 30 40 50 60 70 80 90 100


Output


675105648

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
namespace my_std{
	using namespace std;
	#define pii pair<int,int>
	#define fir first
	#define sec second
	#define MP make_pair
	#define rep(i,x,y) for (int i=(x);i<=(y);i++)
	#define drep(i,x,y) for (int i=(x);i>=(y);i--)
	#define go(x) for (int i=head[x];i;i=edge[i].nxt)
	#define templ template<typename T>
	#define sz 8333
	#define mod 998244353ll
	typedef long long ll;
	typedef double db;
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
	templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
	templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
	templ inline void read(T& t)
	{
		t=0;char f=0,ch=getchar();double d=0.1;
		while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
		while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
		if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
		t=(f?-t:t);
	}
	template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
	char __sr[1<<21],__z[20];int __C=-1,__zz=0;
	inline void Ot(){fwrite(__sr,1,__C+1,stdout),__C=-1;}
	inline void print(int x)
	{
		if(__C>1<<20)Ot();if(x<0)__sr[++__C]='-',x=-x;
		while(__z[++__zz]=x%10+48,x/=10);
		while(__sr[++__C]=__z[__zz],--__zz);__sr[++__C]='\n';
	}
	void file()
	{
		#ifdef NTFOrz
		freopen("a.in","r",stdin);
		#endif
	}
	inline void chktime()
	{
		#ifdef NTFOrz
		cerr<<clock()/1000.0<<'\n';
		#endif
	}
	#ifdef mod
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
	ll inv(ll x){return ksm(x,mod-2);}
	#else
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
	#endif
//	inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;

ll fac[sz],_fac[sz];
void init(){_fac[0]=fac[0]=1;rep(i,1,sz-1) _fac[i]=inv(fac[i]=fac[i-1]*i%mod);}

int r[sz],limit;
void NTT_init(int n)
{
	int l=-1;
	for (limit=1;limit<=n;limit<<=1) ++l;
	rep(i,0,limit-1) r[i]=(r[i>>1]>>1)|((i&1)<<l);
}
void NTT(ll *a,int type)
{
	rep(i,0,limit-1) if (i<r[i]) swap(a[i],a[r[i]]);
	for (int mid=1;mid<limit;mid<<=1)
	{
		ll Wn=ksm(3,mod/mid>>1); if (type==-1) Wn=inv(Wn);
		for (int j=0;j<limit;j+=mid<<1)
		{
			ll w=1;
			rep(k,0,mid-1)
			{
				ll x=a[j+k],y=a[j+k+mid]*w%mod;
				a[j+k]=(x+y)%mod,a[j+k+mid]=(x-y+mod)%mod;
				w=w*Wn%mod;
			}
		}
	}
	if (type==1) return;
	ll I=inv(limit);
	rep(i,0,limit-1) a[i]=a[i]*I%mod;
}

int n,T;
int l[sz],L;

ll dp[55][55][sz],tmp1[sz],tmp2[sz],tmp3[sz];

int main()
{
	file();
	init();
	read(n,T); rep(i,1,n) read(l[i]),L+=l[i]; sort(l+1,l+n+1),reverse(l+1,l+n+1);
	dp[0][0][0]=1; NTT_init(L*2+5);
	rep(i,1,n) rep(t,0,i)
	{
		rep(k,0,L) tmp1[k]=dp[i-1][t][k]; NTT(tmp1,1);
		rep(k,0,(l[i]-1)/T)
		{
			ll p=(l[i]-k*T)*inv(L)%mod;
			tmp2[k]=((k&1)?mod-1:1)*_fac[k]%mod*ksm(p,k)%mod;
			if (k) tmp3[k]=((k&1)?mod-1:1)*_fac[k-1]%mod*ksm(p,k-1)%mod;
		}
		NTT(tmp2,1),NTT(tmp3,1);
		rep(i,0,limit-1) tmp2[i]=tmp2[i]*tmp1[i]%mod,tmp3[i]=tmp3[i]*tmp1[i]%mod;
		NTT(tmp2,-1),NTT(tmp3,-1);
		rep(k,0,L) (dp[i][t][k]+=tmp2[k])%=mod,(dp[i][t+1][k]+=tmp3[k])%=mod;
		rep(k,0,limit-1) tmp1[k]=tmp2[k]=tmp3[k]=0;
	}
//	rep(j,0,L/T) rep(k,0,(l[i]-1)/T) 
//	{
//		ll p=(l[i]-k*T)*inv(L)%mod;
//		(dp[i][j+k][t]+=((k&1)?mod-1:1)*_fac[k]%mod*ksm(p,k)%mod*dp[i-1][j][t]%mod)%=mod;
//		if (k) (dp[i][j+k][t+1]+=((k&1)?mod-1:1)*_fac[k-1]%mod*ksm(p,k-1)%mod*dp[i-1][j][t]%mod)%=mod;
//	}
	ll ans=0;
	rep(j,1,L/T) rep(t,0,min(j,n))
	{
		ll p=1-j*T%mod*inv(L)%mod;
		(ans+=dp[n][t][j]*ksm(1-p,mod-1-(j-t+1))%mod*fac[j-t]%mod+mod)%=mod;
	}
	chktime();
	ans=(mod-ans)%mod;
	cout<<ans;
	return 0;
}