#include<iostream>//6分答案，样例全过，好！ 
#include<cstdio>
#include<cstring>
#define F(i,j,k) for(int i=(j);i<=(k);i++)
#define N 1002
#define P 998244353
using namespace std;
int n,m,T,id,a1;
bool map[N][N];//思路：若第y1到第y2有通路，结果为x1可延长的数量加x2可延长的数量
int mapk[N][N];
bool c,f;
int ans;
void doc();
void dof();
int main(){
	
	cin>>T>>id;
	int k,tmp,tmp2;
	while(T--){
		memset(map,0,sizeof(map));
		memset(mapk,0,sizeof(mapk));
		cin>>n>>m>>c>>f;getchar();
		F(i,1,n){
			F(j,1,m)
			{
				map[i][j]=getchar()=='0'?false:true;
			}
			getchar();
		}
		F(i,1,n)
		F(j,1,m-1){
			if(map[i][j])continue;
			tmp2=j;
			F(h,1,m-j+1)//边界：m-j+1
			{
				if(map[i][tmp2])break;
				else tmp2++;
			}
				a1=tmp2-j>0?tmp2-j-1:0;
				mapk[i][j]=a1;
		}
		doc();
		cout<<' ';
		dof();
		cout<<endl;
	}
	return 0;
}
void doc(){
	int tt,p,anstmp;
	ans=0;
	F(i,1,n-2)
		F(j,1,m-1)
		{
			anstmp=0;
			if(mapk[i][j]){
				tt=mapk[i][j],p=i+1;
				if(map[p][j])p=n+100;//跳出循环 
				while(p<n){
					p++;
					if(!map[p][j])anstmp+=mapk[p][j];
					else break;
				}
			}
			ans+=(anstmp*tt)%P;
			ans%=P;
			
		}
	cout<<ans*c;
}
void dof(){
	int tt,p,anstmp;
	ans=0;
	F(i,1,n-3)
		F(j,1,m-1)
		{
			anstmp=0;
			if(mapk[i][j]){
				tt=mapk[i][j],p=i;
				while(!map[p+1][j]&&p<n){
					p++;
				}//[i,p]
				if(i+2<p){
					F(h,i+2,p-1){
						anstmp+=mapk[h][j]*(p-i-2);
					} 
				}
			}
			ans+=(anstmp*tt)%P;
			ans%=P;
			
		}
	cout<<ans*f;
}