/*
状态压缩类DP + 计数
*/
// include file
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <ctime>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <bitset>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <list>
#include <functional>
using namespace std;
// typedef
typedef long long LL;
typedef unsigned long long ULL;
//
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#define FORi(a,b,c) for(int i=(a);i<(b);i+=c)
#define FORj(a,b,c) for(int j=(a);j<(b);j+=c)
#define FORk(a,b,c) for(int k=(a);k<(b);k+=c)
#define FORp(a,b,c) for(int p=(a);p<(b);p+=c)
#define FORii(a,b,c) for(int ii=(a);ii<(b);ii+=c)
#define FORjj(a,b,c) for(int jj=(a);jj<(b);jj+=c)
#define FORkk(a,b,c) for(int kk=(a);kk<(b);kk+=c)
#define FF(i,a) for(int i=0;i<(a);i++)
#define FFD(i,a) for(int i=(a)-1;i>=0;i--)
#define Z(a) (a<<1)
#define Y(a) (a>>1)
const double eps = 1e-6;
const double INFf = 1e10;
const int INFi = 1000000000;
const double Pi = acos(-1.0);
template<class T> inline T sqr(T a){return a*a;}
template<class T> inline T TMAX(T x,T y)
{
if(x>y) return x;
return y;
}
template<class T> inline T TMIN(T x,T y)
{
if(x<y) return x;
return y;
}
template<class T> inline void SWAP(T &x,T &y)
{
T t = x;
x = y;
y = t;
}
template<class T> inline T MMAX(T x,T y,T z)
{
return TMAX(TMAX(x,y),z);
}
// code begin
// dp[i][j] = sigma(dp[i-1][k]) ,所有j和k可以并存的情况下
//
#define mod 100000000
int N,M;
int mp[13][13];
int pos[13];
int stk[13][4100];
int top[13];
int dp[13][4100];
void Init()
{
memset(top,0,sizeof(top));
int t;
FORi(1,N+1,1)
{
FORj(0,1<<M,1)
{
t = j;
// 判非法
if( (t<<1)&j ) continue;
//
if( t&pos[i]) continue;
stk[i][top[i]++] = t;
}
}
stk[0][top[0]++] = 0;
}
int main()
{
read;
write;
while(scanf("%d %d",&N,&M)!=-1)
{
FORi(1,N+1,1)
{
pos[i] = 0;
FORj(0,M,1)
{
scanf("%d",&mp[i][j]);
pos[i] = pos[i]*2+1-mp[i][j];
}
}
Init();
memset(dp,0,sizeof(dp));
dp[0][0] = 1;
FORi(1,N+1,1)
{
//枚举当前状态
FORj(0,top[i],1)
{
//枚举之前状态
FORk(0,top[i-1],1)
{
if( stk[i][j]&stk[i-1][k] ) continue;
dp[i][j] += dp[i-1][k];
if(dp[i][j]>=mod)
dp[i][j] -= mod;
}
}
}
int ans = 0;
FORi(0,top[N],1)
{
ans += dp[N][i];
if(ans>=mod)
ans-=mod;
}
printf("%d\n",ans);
}
return 0;
}