/*
    D[i][j] = in cate moduri pot pune exact j obiecte in i multimi
    D[i][j] = D[i-1][0] + D[i-1][1] + ... + D[i-1][j]

    generalizare:

    D[i][j] = in cate moduri pot pune cel mult j obiecte in i multimi
    D[i][j] = D[i-1][j] + D[i][j-1]
*/
#include <cstdio>
#include <algorithm>

#define DIM1 33
#define DIM2 10300
#define MOD 3210121
using namespace std;

int N, K, S, sol, sum, Current[DIM1];
int D[DIM1][DIM2], A[DIM1][DIM1], B[DIM1][DIM1];

void backTrack (int lev, int val) {

    if (lev == N) { /* ok am atins o configuratie corecta, vad daca o scad/adun */
        if (!val) return;
        int sum = 0;

        for (int i = 1; i <= K; i ++)
            sum += Current[i];

        if (sum > S) return;

        if (val%2) sol = (sol + D[S-sum][K+1] + MOD) % MOD;
        else       sol = (sol - D[S-sum][K+1] + MOD) % MOD;
    } else { /* configuratia nu este buna, incerc sa o imbunatatesc */

        backTrack (lev + 1, val + 0); /* nu incerc sa adaug o reteta */

        for (int i = 1; i <= K; i ++) {
            B[lev][i]  = Current[i];
            Current[i] = max (Current[i], A[lev][i]);
        }

        backTrack (lev + 1, val + 1); /* incerc sa adaug o reteta */

        for (int i = 1; i <= K; i ++)
            Current[i] = B[lev][i];
    }
    return;
}

int main () {

    freopen ("cowfood.in" ,"r", stdin );
    freopen ("cowfood.out","w", stdout);

    scanf ("%d %d %d", &K, &S, &N);

    for (int i = 0; i < N; i ++)
        for (int j = 1; j <= K; j ++)
            scanf ("%d", &A[i][j]);

    for (int j = 0; j <= K + 1; j ++)
        D[0][j] = 1;

    for (int i = 1; i <= S + 1; i ++) {
        for (int j = 0; j <= K + 1; j ++) {
            if (j == 1 || !j) D[i][j] = 1; else
            if (i == 1)       D[i][j] = j;
            else D[i][j] = (D[i-1][j] + D[i][j-1]) % MOD;
        }
    }

    backTrack (0, 0);
    sol = (D[S][K+1] - sol + MOD) % MOD;
    sol = (sol - S*K - 1 + MOD) % MOD;

    printf ("%d\n", sol);

    return 0;
}