#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

//template
#define rep(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define ALL(v) (v).begin(),(v).end()
typedef long long int ll;
const int inf = 0x3fffffff; const ll INF = 0x1fffffffffffffff; const double eps=1e-12;
template<typename T>inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<typename T>inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
//end

template<unsigned mod=998244353>struct fp {
   unsigned v;
   static unsigned get_mod(){return mod;}
   unsigned inv() const{
      int tmp,a=v,b=mod,x=1,y=0;
      while(b)tmp=a/b,a-=tmp*b,swap(a,b),x-=tmp*y,swap(x,y);
      if(x<0){x+=mod;} return x;
   }
   fp():v(0){}
   fp(ll x):v(x>=0?x%mod:mod+(x%mod)){}
   fp operator-()const{return fp(-v);}
   fp pow(ll t){fp res=1,b=*this; while(t){if(t&1)res*=b;b*=b;t>>=1;} return res;}
   fp& operator+=(const fp& x){if((v+=x.v)>=mod)v-=mod; return *this;}
   fp& operator-=(const fp& x){if((v+=mod-x.v)>=mod)v-=mod; return *this;}
   fp& operator*=(const fp& x){v=ll(v)*x.v%mod; return *this;}
   fp& operator/=(const fp& x){v=ll(v)*x.inv()%mod; return *this;}
   fp operator+(const fp& x)const{return fp(*this)+=x;}
   fp operator-(const fp& x)const{return fp(*this)-=x;}
   fp operator*(const fp& x)const{return fp(*this)*=x;}
   fp operator/(const fp& x)const{return fp(*this)/=x;}
   bool operator==(const fp& x)const{return v==x.v;}
   bool operator!=(const fp& x)const{return v!=x.v;}
}; using Fp=fp<>;

//※TLE解法
//「山」は一番後ろの要素からdequeみたいに詰めるゲームと言い換えられる
//片方は必ず直前に詰めた要素なのでもう一方を持てばよい

Fp from[3010][3010],to[3010][3010];
//dp[i][j]... i:もう片方の要素idx j:危険度 = 通り数

int main(){
   int n; cin>>n;
   vector<int> a(n); rep(i,0,n)cin>>a[i];
   reverse(ALL(a));
   from[0][0]=1;
   rep(k,1,n){
      rep(i,0,n)rep(j,0,3010)to[i][j]=0;
      rep(i,0,k)rep(j,0,3010)if(from[i][j]!=0){
         if(j>a[k-1]-a[k]) to[i][j]+=from[i][j];
         else to[i][a[k-1]-a[k]]+=from[i][j];
         if(j>a[i]-a[k])to[k-1][j]+=from[i][j];
         else to[k-1][a[i]-a[k]]+=from[i][j];
      } swap(from,to);
   }
   Fp res;
   rep(i,0,n)rep(j,0,3010)res+=from[i][j]*j;
   cout<<res.v<<endl;
   return 0;
}