//判断每个气球需要多少次才能扎破
//大于k的，那么就不满足条件
//可以看作最小点覆盖问题
//分为两个集合，横坐标集合，纵坐标集合
//然后每个点的坐标在这两个集合中且两点之间有连线
//然后就是求最小点覆盖
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn = 105;
int n, k;
int g[maxn][maxn];
int color[52];
bool st[maxn];
int match[maxn];
int nowcolor;
vector<int>v;
void init() {
memset(g, 0, sizeof(g));
memset(color, 0, sizeof(color));
v.clear();
}
bool find(int x) {
for (int i = 1; i <= n; ++i) {
if (g[x][i] == nowcolor && st[i] == 0) {
st[i] = 1;
if (match[i] == -1 || find(match[i])) {
match[i] = x;
return true;
}
}
}
return false;
}
int solve(int color) {
nowcolor = color;
memset(match, -1, sizeof(match));
int ans = 0;
for(int i = 1; i <= n; ++i) {
memset(st, 0, sizeof(st));
if (find(i))
ans++;
}
return ans;
}
int main() {
while(~scanf("%d%d", &n, &k) && n && k) {
init();
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
{
scanf("%d", &g[i][j]);
color[g[i][j]] = 1;
}
for(int i = 1; i <= 50; ++i)
if (color[i] == 1)
if(solve(i) > k)
v.push_back(i);
if (v.size() == 0)
printf("-1");
else {
for(int i = 0; i < v.size(); ++i) {
if(i)
printf(" ");
printf("%d", v[i]);
}
}
printf("
");
}
return 0;
}