The meaning of the question: give you an n*m matrix, 0 means no one can stand, 1 means no one can stand, and no one can stand between adjacent grids (up, down, left, right, four directions). Ask how many ways to stand.
Consider each row, facilitate each state, and maintain the state of the next row based on the current row, complexity o(1<<m)^2*o(n)
The adjacent left and right can be judged by j&(j>>1)
The adjacent upper and lower can be judged by k&j
/// .-~~~~~~~~~-._ _.-~~~~~~~~~-.
/// __.' ~. .~ `.__
/// .'// \./ \\`.
/// .'// | \\`.
/// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`.
/// .'//.-" `-. | .-' "-.\\`.
/// .'//______.============-.. \ | / ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma GCC optimize("Ofast")
#pragma comment(linker, "/STACK:102400000,102400000")
#pragma GCC target(sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx)
#include<bits/stdc++.h>
using namespace std;
#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
//#define ll long long
#define mp make_pair
#define pb push_back
#define sz size
#define fi first
#define se second
#define pf printf
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e2+7;
const int _=1e5+10;
const double EPS=1e-8;
const double eps=1e-8;
const LL mod=100000000;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}
inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}
void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//cerr << "run time is " << clock() << endl;
int n,m;
int a[_];
int dp[15][1<<13];
void solve() {
W(s_2(n,m)!=EOF) {
me(dp,0);
me(a,0);
dp[0][0]=1;
FOR(1,n,i)
FOr(0,m,j) {
int x;
s_1(x);
a[i]|=(x<<j);
}
int sta=1<<m;
FOR(1,n,i) {
for(int j=0;j<sta;j++) {
if((j&a[i])!=j) continue;
if(j&(j>>1)) continue;
for(int k=0;k<sta;k++) {
if(k&j) continue;
dp[i][j]=(dp[i][j]+dp[i-1][k])%mod;
}
}
}
int ans=0;
for(int i=0;i<sta;i++)
ans=(ans+dp[n][i])%mod;
print(ans);
}
}
int main() {
//freopen( "1.in" , "r" , stdin );
//freopen( "1.out" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++) {
//printf("Case %d:\n",cas);
solve();
}
}