/*
    題意：n*m的棋盤，要放中國象棋的炮，問有多少種合法的放法（空也算一種）
           對于炮，只要沒有3個或以上的在同一行或同一列就算合法了
    sample:
        1*3 的棋盤有7種
        不放 算1種
        放1個炮的有3種
        放2個炮的有3種

    可以知道每一行、一列最多只能放2個炮。即可以放0,1,2個
    通過記錄之前那些放0個炮，1個炮，2個炮的列的數目即可！！
    dp[n,x,y]表示在前n行中，放0個炮的列數為x,放1個炮的列數為y 的方案數
    轉移即在該行 不放，放1個，放2個 
    不放
    dp[n,x,y]   ->dp[n+1,x,y]

    放1個，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個，2個都放在x，2個都放在x，1個x 1個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]

    從當前狀態推到后繼狀態會容易理解一點

    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 炮個數為0的列數  炮個數為1的列數

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;
}