/*
题意：k个挤奶机，每天可以给m头奶牛挤奶，让奶牛到达挤奶机的最大距离最小
题解：很明朗的二分答案的题目，不解释
先floyd求任意点之间的最短路，然后二分答案建图判断是否满足要求
*/
#include <cstdio>
#include <iostream>
#include<queue>
#include<set>
#include<ctime>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<cstring>
using namespace std;
const int inf=1<<30;
const int maxn=1009;
int dis[maxn][maxn];
struct edge
{
int v, next;
int val;
} net[ 500100 ];
int k,c,m;
int level[maxn], Qu[maxn], out[maxn],next[maxn];
class Dinic {
public:
int end;
Dinic() {
end = 0;
memset( next, -1, sizeof(next) );
}
inline void insert( int x, int y, int c) {
net[end].v = y, net[end].val = c,
net[end].next = next[x],
next[x] = end ++;
net[end].v = x, net[end].val = 0,
net[end].next = next[y],
next[y] = end ++;
}
bool BFS( int S, int E ) {
memset( level, -1, sizeof(level) );
int low = 0, high = 1;
Qu[0] = S, level[S] = 0;
for( ; low < high; ) {
int x = Qu[low];
for( int i = next[x]; i != -1; i = net[i].next ) {
if( net[i].val == 0 ) continue;
int y = net[i].v;
if( level[y] == -1 ) {
level[y] = level[x] + 1;
Qu[ high ++] = y;
}
}
low ++;
}
return level[E] != -1;
}
int MaxFlow( int S, int E ){
int maxflow = 0;
for( ; BFS(S, E) ; ) {
memcpy( out, next, sizeof(out) );
int now = -1;
for( ;; ) {
if( now < 0 ) {
int cur = out[S];
for(; cur != -1 ; cur = net[cur].next )
if( net[cur].val && out[net[cur].v] != -1 && level[net[cur].v] == 1 )
break;
if( cur >= 0 ) Qu[ ++now ] = cur, out[S] = net[cur].next;
else break;
}
int u = net[ Qu[now] ].v;
if( u == E ) {
int flow = inf;
int index = -1;
for( int i = 0; i <= now; i ++ ) {
if( flow > net[ Qu[i] ].val )
flow = net[ Qu[i] ].val, index = i;
}
maxflow += flow;
for( int i = 0; i <= now; i ++ )
net[Qu[i]].val -= flow, net[Qu[i]^1].val += flow;
for( int i = 0; i <= now; i ++ ) {
if( net[ Qu[i] ].val == 0 ) {
now = index - 1;
break;
}
}
}
else{
int cur = out[u];
for(; cur != -1; cur = net[cur].next )
if (net[cur].val && out[net[cur].v] != -1 && level[u] + 1 == level[net[cur].v])
break;
if( cur != -1 )
Qu[++ now] = cur, out[u] = net[cur].next;
else out[u] = -1, now --;
}
}
}
return maxflow;
}
};
void floyd()
{
for(int g=1;g<=k+c;g++)
{
for(int i=1;i<=k+c;i++)if(g!=i&&dis[i][g]!=-1)
{
for(int j=1;j<=k+c;j++)if(g!=j&&dis[g][j]!=-1)
{
if(dis[i][j]==-1||dis[i][j]>dis[i][g]+dis[g][j])
dis[i][j]=dis[i][g]+dis[g][j];
}
}
}
}
void build(int lim,Dinic &my)
{
for(int i=1;i<=k;i++)
my.insert(0,i,m);
for(int i=1;i<=k;i++)
{
for(int j=k+1;j<=k+c;j++)
if(dis[i][j]<=lim&&dis[i][j]!=-1)
{
my.insert(i,j,1);
}
}
for(int j=k+1;j<=k+c;j++)
my.insert(j,k+c+1,1);
}
void solve()
{
int low=0,hei=20000,mid;
while(low<hei-1)
{
mid=(low+hei)/2;
Dinic my;
build(mid,my);
if(my.MaxFlow(0,k+c+1)==c)
hei=mid;
else
low=mid;
}
printf("%d\n",hei);
}
int main()
{
while(scanf("%d%d%d",&k,&c,&m)!=EOF)
{
for(int i=1;i<=k+c;i++)
for(int j=1;j<=k+c;j++)
{
scanf("%d",&dis[i][j]);
if(i!=j&&dis[i][j]==0)
dis[i][j]=-1;
}
floyd();
solve();
}
}
