//二分枚举最大距离的最小值+Floyd找到最短路+Dinic算法

//参考图论算法书，并对BFS构建层次网络算法进行改进

//Time:157Ms    Memory:652K

#include<iostream>

#include<cstring>

#include<cstdio>

#include<algorithm>

#include<queue>

using namespace std;

#define MAX 250

#define INF 0x3f3f3f3f

int K, C, M;

int s, t;

int d[MAX][MAX];	//各点间最短距离

int res[MAX][MAX];	//残留网络

int lev[MAX];

void build_map(int limit)

{

    memset(res,0,sizeof(res));

	for (int i = K + 1; i <= K + C; i++)

		res[s][i] = 1;

	for (int i = 1; i <= K; i++)

		res[i][t] = M;

	for (int i = K + 1; i <= K + C; i++)

		for (int j = 1; j <= K; j++)

			if (d[i][j] <= limit)	res[i][j] = 1;

}

bool bfs()  //BFS标记层次网络

{

    memset(lev, -1, sizeof(lev));

	queue<int> q;

	q.push(s);

	lev[s] = 0;

	while (!q.empty()) {    //构建层次网络

		int cur = q.front();

		q.pop();

		for(int i = 1; i <= t; i++)

        {

            if(lev[i] == -1 && res[cur][i])    //未访问且正向有流量

            {

                q.push(i);

                lev[i] = lev[cur] + 1;

            }

        }

	}

	return lev[t] != -1;

}

int dfs(int v, int alpha)  //DFS进行多次增广

{

    if(v == t || alpha == 0)  return alpha;

    int src = alpha;    //原可改进量

    for(int i = 1; i <= t; i++)

    {

        if(res[v][i] && lev[i] == lev[v] + 1){  //识别下一层次

            int tmp = dfs(i, min(alpha, res[v][i]));

            res[v][i] -= tmp;

            res[i][v] += tmp;

            alpha -= tmp;   //可改进量减少

        }

    }

    return src - alpha; //总改进量

}

int main()

{

	//freopen("in.txt", "r", stdin);

	scanf("%d%d%d", &K,&C,&M);

	s = 0;  t = K + C + 1;  //源点-汇点

	for (int i = 1; i < t; i++)

		for (int j = 1; j < t; j++)

		{

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

			if (d[i][j] == 0)	d[i][j] = INF;

		}

	for (int k = 1; k < t; k++)

		for (int i = 1; i < t; i++)

		{

			if (d[i][k] != INF) {

				for (int j = 1; j < t; j++)

					d[i][j] = min(d[i][j], d[i][k] + d[k][j]);

			}

		}

	s = 0;	t = K + C + 1;	//源点 汇点

	int l = 0, r = 9000;

	while (l < r)

	{

		int ans = 0;    //到达目的地的奶牛数量

		int mid = (l + r) / 2;

		build_map(mid);

		while (bfs())

			ans += dfs(0,INF); //第二参数指定该点可改进量

        ans == C ? r = mid: l = mid+1;

	}

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

	return 0;

}