/*
    题意：n*m的棋盘，要放中国象棋的炮，问有多少种合法的放法（空也算一种）
           对于炮，只要没有3个或以上的在同一行或同一列就算合法了
    sample:
        1*3 的棋盘有7种
        不放 算1种
        放1个炮的有3种
        放2个炮的有3种

    可以知道每一行、一列最多只能放2个炮。即可以放0,1,2个
    通过记录之前那些放0个炮，1个炮，2个炮的列的数目即可！！
    dp[n,x,y]表示在前n行中，放0个炮的列数为x,放1个炮的列数为y 的方案数
    转移即在该行 不放，放1个，放2个 
    不放
    dp[n,x,y]   ->dp[n+1,x,y]

    放1个，可放在x，也可放在y
    dp[n,x,y]  *x ->dp[n+1,x-1,y+1]
    dp[n,x,y]  *y ->dp[n+1,x,y-1]

    放2个，2个都放在x，2个都放在x，1个x 1个y
    dp[n,x,y]  *c(x,2) ->dp[n+1,x-2,y+2]
    dp[n,x,y]  *c(y,2) ->dp[n+1,x,y-2]
    dp[n,x,y]  *x*y    ->dp[n+1,x-1,y]

    从当前状态推到后继状态会容易理解一点

    http://61.187.179.132:8080/JudgeOnline/showproblem?problem_id=1801
*/
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn = 105;
const int mod = 9999973;

int dp[2][maxn][maxn];//前i行 炮个数为0的列数  炮个数为1的列数

int main()
{

    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        if(n<m)swap(n,m);
        int now = 1,next = 0;
        memset(dp[next],0,sizeof(dp[next]));
        dp[next][m][0]=1;//
        for(int i=0;i<n;i++)
        {
            swap(now,next);
            memset(dp[next],0,sizeof(dp[next]));            
            for(int a0=0;a0<=m;a0++)
                for(int a1=0;a0+a1<=m;a1++)
                    if(dp[now][a0][a1])
                    {
                        int t=dp[now][a0][a1];

                        //not put
                        dp[next][a0][a1]+=t;
                        dp[next][a0][a1]%=mod;

                        //put one
                        if(a0)
                        {
                            dp[next][a0-1][a1+1]+=(long long)t*a0%mod;
                            dp[next][a0-1][a1+1]%=mod;
                        }
                        if(a1)
                        {
                            dp[next][a0][a1-1]+=(long long)t*a1%mod;
                            dp[next][a0][a1-1]%=mod;
                        }
                    
                        //put two
                        if(a0>=2)
                        {
                            dp[next][a0-2][a1+2]+=(long long)t*a0*(a0-1)/2%mod;
                            dp[next][a0-2][a1+2]%=mod;
                        }
                        if(a1>=2)
                        {
                            dp[next][a0][a1-2]+=(long long)t*a1*(a1-1)/2%mod;
                            dp[next][a0][a1-2]%=mod;
                        }
                        if(a0&&a1)
                        {
                            dp[next][a0-1][a1]+=(long long)t*a0*a1%mod;
                            dp[next][a0-1][a1]%=mod;
                        }
                    }
        }

        long long ans = 0;
        for(int a0=0;a0<=m;a0++)
            for(int a1=0;a0+a1<=m;a1++)
            {
                ans+=dp[next][a0][a1];
            }
        printf("%lld\n",ans%mod);
    }
    return 0;
}