Problem: You have c_1 letters 'a', c_2 letters 'b', ..., c_{26} letters 'z'. You want to build a beautiful string of length n from them (obviously, you cannot use the i-th letter more than c_i times). Each c_i is greater than n/3.

A string is called beautiful if there are no palindromic contiguous substrings of odd length greater than 1 in it. For example, the string "abacaba" is not beautiful, it has several palindromic substrings of odd length greater than 1 (for example, "aca"). Another example: the string "abcaa" is beautiful.

Calculate the number of different strings you can build, and print the answer modulo 998244353.

Input

The first line contains one integer n (3 ≤ n ≤ 400).

The second line contains 26 integers c_1, c_2, ..., c_{26} (n/3 < c_i ≤ n).

Output

Print one integer — the number of strings you can build, taken modulo 998244353.

Examples

Input


4
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2


Output


422500


Input


3
2 2 2 2 2 2 3 3 3 2 2 2 2 2 2 3 3 3 2 2 3 2 2 3 2 2


Output


16900


Input


400
348 322 247 158 209 134 151 267 268 176 214 379 372 291 388 135 147 304 169 149 193 351 380 368 181 340


Output


287489790

Solution (in CPP):

Solution:
 #include <bits/stdc++.h>
// #include <atcoder/all>

//using namespace atcoder;
//tabaicho  see https://boostjp.github.io/tips/multiprec-int.html   
// #include <boost/multiprecision/cpp_int.hpp>

// using namespace boost::multiprecision;

// cpp_int
// int128_t
// int256_t
// int512_t
// int1024_t

// uint128_t
// uint256_t
// uint512_t
// uint1024_t



#define int long long
 #define inf  1000000007
// #define inf  998244353

 #define pa pair<int,int>
 #define ll long long
 #define PI 3.14159265358979323846
 #define  mp make_pair
 #define  pb push_back
 #define EPS (1e-8)
 
      using namespace std;
                                          
 int dx[8]={0,1,0,-1,1,1,-1,-1};
 int dy[8]={1,0,-1,0,-1,1,1,-1};
                                            
class pa3{
	public:
	int x;
	int y,z;
	pa3(int x=0,int y=0,int z=0):x(x),y(y),z(z) {}
	bool operator < (const pa3 &p) const{
		if(x!=p.x) return x<p.x;
		if(y!=p.y) return y<p.y;
		 return z<p.z;
		//return x != p.x ? x<p.x: y<p.y;
	}
	bool operator > (const pa3 &p) const{
		if(x!=p.x) return x>p.x;
		if(y!=p.y) return y>p.y;
		 return z>p.z;
		//return x != p.x ? x<p.x: y<p.y;
	}
	bool operator == (const pa3 &p) const{
		return x==p.x && y==p.y && z==p.z;
	}
		bool operator != (const pa3 &p) const{
			return !( x==p.x && y==p.y && z==p.z);
	}

};

class pa4{
	public:
	int x;
	int y,z,w;
	pa4(int x=0,int y=0,int z=0,int w=0):x(x),y(y),z(z),w(w) {}
	bool operator < (const pa4 &p) const{
		if(x!=p.x) return x<p.x;
		if(y!=p.y) return y<p.y;
		if(z!=p.z)return z<p.z;
		return w<p.w;
		//return x != p.x ? x<p.x: y<p.y;
	}
	bool operator > (const pa4 &p) const{
		if(x!=p.x) return x>p.x;
		if(y!=p.y) return y>p.y;
		if(z!=p.z)return z>p.z;
		return w>p.w;
		//return x != p.x ? x<p.x: y<p.y;
	}
	bool operator == (const pa4 &p) const{
		return x==p.x && y==p.y && z==p.z &&w==p.w;
	}
		

};
class pa2{
	public:
	int x,y;
	pa2(int x=0,int y=0):x(x),y(y) {}
	pa2 operator + (pa2 p) {return pa2(x+p.x,y+p.y);}
	pa2 operator - (pa2 p) {return pa2(x-p.x,y-p.y);}
	bool operator < (const pa2 &p) const{
		return y != p.y ? y<p.y: x<p.x;
	}
	bool operator > (const pa2 &p) const{
		return x != p.x ? x<p.x: y<p.y;
	}
	bool operator == (const pa2 &p) const{
		return abs(x-p.x)==0 && abs(y-p.y)==0;
	}
	bool operator != (const pa2 &p) const{
		return !(abs(x-p.x)==0 && abs(y-p.y)==0);
	}
		

};



string itos( int i ) {
	ostringstream s ;
	s << i ;
	return s.str() ;
}
 
int Gcd(int v,int b){
	if(v==0) return b;
	if(b==0) return v;
	if(v>b) return Gcd(b,v);
	if(v==b) return b;
	if(b%v==0) return v;
	return Gcd(v,b%v);
}


 int mod;
int extgcd(int a, int b, int &x, int &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    int d = extgcd(b, a%b, y, x);
    y -= a/b * x;
    return d;
}
pa operator+(const pa & l,const pa & r) {   
    return {l.first+r.first,l.second+r.second};                                    
}    
pa operator-(const pa & l,const pa & r) {   
    return {l.first-r.first,l.second-r.second};                                    
}  

ostream& operator<<(ostream& os, const vector<int>& VEC){
	for(auto v:VEC)os<<v<<" ";
    return os;
}

ostream& operator<<(ostream& os, const pa& PAI){
	os<<PAI.first<<" : "<<PAI.second;
    return os;
}

ostream& operator<<(ostream& os, const pa3& PAI){
	os<<PAI.x<<" : "<<PAI.y<<" : "<<PAI.z;
    return os;
}

ostream& operator<<(ostream& os, const pa4& PAI){
	os<<PAI.x<<" : "<<PAI.y<<" : "<<PAI.z<<" : "<<PAI.w;
    return os;
}

ostream& operator<<(ostream& os, const vector<pa>& VEC){
	for(auto v:VEC)os<<v<<" ";
    return os;
}


int beki(int wa,int rr,int warukazu){
	if(rr==0) return 1%warukazu;
	if(rr==1) return wa%warukazu;
	wa%=warukazu;
	if(rr%2==1) return ((ll)beki(wa,rr-1,warukazu)*(ll)wa)%warukazu;
	ll zx=beki(wa,rr/2,warukazu);
	return (zx*zx)%warukazu;
}

              
int pr[2100000];
int inv[2100000];



int comb(int nn,int rr){
	if(rr<0 || rr>nn || nn<0) return 0;
	int r=pr[nn]*inv[rr];
	r%=mod;
	r*=inv[nn-rr];
	r%=mod;
	return r;
}

void gya(int ert){
	pr[0]=1;
	for(int i=1;i<=ert;i++){
		pr[i]=((ll)pr[i-1]*i)%mod;
	}
		inv[ert]=beki(pr[ert],mod-2,mod);
	for(int i=ert-1;i>=0;i--){
		inv[i]=(ll)inv[i+1]*(i+1)%mod;
	}
}


                


                
              //   cin.tie(0);
    		//	ios::sync_with_stdio(false);
    			//priority_queue<pa3,vector<pa3>,greater<pa3>> pq;            
                 //sort(ve.begin(),ve.end(),greater<int>());
//    mt19937(clock_per_sec);
  //  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()) ;
int c[26];


int dp1[410][410][2]={};

void mae1(int n,int m){
	dp1[1][0][0]=1;
	dp1[1][1][1]=1;
	
	for(int i=2;i<=n;i++){
		for(int j=0;j<=i;j++){
			dp1[i][j][0]=(dp1[i-1][j][1]+dp1[i-1][j][0]*24)%mod;
			if(j)dp1[i][j][1]=(dp1[i-1][j-1][0]*25)%mod;
			
		}
	}
	for(int i=n+1;i<=n+1;i++){
		for(int j=0;j<=i;j++){
			dp1[i][j][0]=(dp1[i-1][j][1]+dp1[i-1][j][0]*25)%mod;
			if(j)dp1[i][j][1]=(dp1[i-1][j-1][0]*25+dp1[i-1][j-1][1])%mod;
			
		}
	}
	for(int i=n+2;i<=m+n;i++){
		for(int j=0;j<=i;j++){
			dp1[i][j][0]=(dp1[i-1][j][1]+dp1[i-1][j][0]*24)%mod;
			if(j)dp1[i][j][1]=(dp1[i-1][j-1][0]*25)%mod;
			
		}
	}
	
}

int sol1(int x,int n){
	int ans=0;
	for(int i=c[x]+1;i<=n;i++)ans+=dp1[n][i][0]*25+dp1[n][i][1];

	return ans%mod;
}


int dp2[410][220][220][3]={};

int ue[410][410]={};
int J=0;
void mae2(int n,int m){
	dp2[1][0][0][0]=1;
	dp2[1][1][0][1]=1;
	dp2[1][0][1][2]=1;
	
	
	for(int i=2;i<=n;i++){
		//for(int j=0;j<=i;j++)for(int k=0;k<=i;k++){
		for(int j=0;j<=(i+15)/2;j++)for(int k=0;k<=(i+15)/2;k++){
			dp2[i][j][k][0]=dp2[i-1][j][k][0]*23+dp2[i-1][j][k][1]+dp2[i-1][j][k][2];
			if(j)dp2[i][j][k][1]=dp2[i-1][j-1][k][0]*24+dp2[i-1][j-1][k][2];
			if(k)dp2[i][j][k][2]=dp2[i-1][j][k-1][0]*24+dp2[i-1][j][k-1][1];
			dp2[i][j][k][0]%=mod;
			dp2[i][j][k][1]%=mod;
			dp2[i][j][k][2]%=mod;
			
		}
	}
	for(int i=n+1;i<=n+1;i++){
		//for(int j=0;j<=i;j++)for(int k=0;k<=i;k++){
		for(int j=0;j<=(i+15)/2;j++)for(int k=0;k<=(i+15)/2;k++){
			dp2[i][j][k][0]=dp2[i-1][j][k][0]*24+dp2[i-1][j][k][1]+dp2[i-1][j][k][2];
			if(j)dp2[i][j][k][1]=dp2[i-1][j-1][k][0]*24+dp2[i-1][j-1][k][1]+dp2[i-1][j-1][k][2];
			if(k)dp2[i][j][k][2]=dp2[i-1][j][k-1][0]*24+dp2[i-1][j][k-1][1]+dp2[i-1][j][k-1][2];
			dp2[i][j][k][0]%=mod;
			dp2[i][j][k][1]%=mod;
			dp2[i][j][k][2]%=mod;
			
		}
	}
	for(int i=n+2;i<=n+m;i++){
		//for(int j=0;j<=i;j++)for(int k=0;k<=i;k++){
		for(int j=0;j<=(i+15)/2;j++)for(int k=0;k<=(i+15)/2;k++){
			dp2[i][j][k][0]=dp2[i-1][j][k][0]*23+dp2[i-1][j][k][1]+dp2[i-1][j][k][2];
			if(j)dp2[i][j][k][1]=dp2[i-1][j-1][k][0]*24+dp2[i-1][j-1][k][2];
			if(k)dp2[i][j][k][2]=dp2[i-1][j][k-1][0]*24+dp2[i-1][j][k-1][1];
			dp2[i][j][k][0]%=mod;
			dp2[i][j][k][1]%=mod;
			dp2[i][j][k][2]%=mod;
			
		}
	}
	
	for(int i=0;i<=207;i++)for(int j=0;j<=207;j++)ue[i][j]=(dp2[n+m][i][j][0]*24+dp2[n+m][i][j][1]+dp2[n+m][i][j][2])%mod;
	
	for(int i=400;i>=0;i--)for(int j=400;j>=0;j--)ue[i][j]=(ue[i+1][j]+ue[i][j+1]-ue[i+1][j+1]+mod+ue[i][j])%mod;
	
	
}
int sol2(int x,int y){

	return ue[c[x]+1][c[y]+1];
}


void solve(){
int n;
	cin>>n;
	mod=998244353;
	for(int i=0;i<26;i++)cin>>c[i];
	//for(int i=0;i<26;i++)c[i]=3;
	int m=n/2;
	n-=m;
	int z0=beki(25,n+m-2,mod)%mod*676%mod;
	mae1(n,m);
	int z1=0;
	for(int i=0;i<26;i++)z1+=sol1(i,n+m);
	z1%=mod;
	mae2(n,m);
	int z2=0;
	for(int i=0;i<26;i++)for(int j=i+1;j<26;j++)z2+=sol2(i,j);
	
	z2%=mod;
	//cout<<z0<<" "<<z1<<" "<<z2<<endl;
	cout<<(z0-z1+z2+mod)%mod<<endl;

}


signed main(){
	mod=inf;
	cin.tie(0);
	ios::sync_with_stdio(false);

int n=1;
//	cin>>n;
for(int i=0;i<n;i++)solve();
	return 0;

	
}