#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cstring>

using namespace std;

const int maxn = ;

const int inf = ;

int n, ans;

int cap[maxn][maxn], pre[maxn];

int cost[maxn][maxn], dis[maxn];

int que[maxn];

bool vis[maxn];

bool spfa()

{

    int i, head = , tail = ;

    for(i = ; i <= n; i ++){

        dis[i] = inf;

        vis[i] = false;

    }

    dis[] = ;

    que[] = ;

    while(tail != head){

        int u = que[head];

        vis[u] = true;

        for(i = ; i <= n; i ++)

            if(cap[u][i] && dis[i] > dis[u] + cost[u][i]){

                dis[i] = dis[u] + cost[u][i];

                pre[i] = u;

                if(!vis[i]){

                    vis[i] = true;

                    que[tail ++] = i;

                    if(tail == maxn) tail = ;

                }

            }

        vis[u] = false;

        head ++;

        if(head == maxn) head = ;

    }

    if(dis[n] == inf) return false;

    return true;

}

void end()

{

    int i, sum = inf;

    for(i = n; i != ; i = pre[i])

        sum = min(sum, cap[pre[i]][i]);

    for(i = n; i != ; i = pre[i]){

        cap[pre[i]][i] -= sum;

        cap[i][pre[i]] += sum;

        ans += cost[pre[i]][i] * sum;

    }

}

int main()

{

    int N, M, K, i, j, k;

    int need[maxn][maxn], NeedK[maxn];

    int have[maxn][maxn], HaveK[maxn];

    while(cin>>N>>M>>K,N,M,K)

    {

        memset(NeedK, , sizeof(NeedK));

        for(i = ; i <= N; i ++)

            for(j = ; j <= K; j ++){

                scanf("%d", &need[i][j]);    //  第i个客户需要第j种货物的量。

                NeedK[j] += need[i][j];      //  第j种货物总共需要的量。

            }

        memset(HaveK, , sizeof(HaveK));

        for(i = ; i <= M; i ++)

            for(j = ; j <= K; j ++){

                scanf("%d", &have[i][j]);    //  第i个仓库存储第j种货物的量。

                HaveK[j] += have[i][j];      //  第j种货物总共需要的量。

            }

        bool flag = true;

        for(i = ; i <= K; i ++)             //  判断所有货物是否足够。

            if(NeedK[i] > HaveK[i]){

                flag = false; break;

            }

        ans = ;

        n = N + M + ;

        for(k = ; k <= K; k ++){

            memset(cap, , sizeof(cap));

            for(i = ; i <= N; i ++)         //  建图。

                for(j = ; j <= M; j ++){

                    scanf("%d", &cost[j][M+i]);

                    cost[M+i][j] = -cost[j][M+i];

                    cap[j][M+i] = inf;

                }

            if(!flag) continue;

            for(i = ; i <= M; i ++){

                cap[][i] = have[i][k];

                cost[][i] = cost[i][] = ;

            }

            for(i = ; i <= N; i ++){

                cap[M+i][n] = need[i][k];

                cost[M+i][n] = cost[n][M+i] = ;

            }

            while(spfa()) end();            //  最小费用最大流算法。

        }

        if(flag) cout << ans << endl;

        else cout << - << endl;

    }

    return ;

}