/*

     题意：就是从上到下，找到最短路，输出路径

     DP+路径：状态转移方程：dp[i][j] = min (dp[i-1][j], dp[i][j-1], dp[i][j+1]) + a[[i][j];    (类似数塔问题)

             关键在记录路径，可以用pre[x][y] = -1/0/1/2 区分，DFS回溯输出

     详细解释：http://www.cnblogs.com/staginner/archive/2012/05/02/2479658.html

 */

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 #include <cmath>

 #include <vector>

 #include <stack>

 using namespace std;

 const int MAXN = 1e2 + ;

 const int MAXM = 5e2 + ;

 const int INF = 0x3f3f3f3f;

 int a[MAXN][MAXM];

 int dp[MAXN][MAXM];

 int pre[MAXN][MAXM];

 void DFS(int x, int y)

 {

     if (pre[x][y] == -)    {printf ("%d", y);    return ;}

     if (pre[x][y] == )        DFS (x-, y);

     else if (pre[x][y] == )    DFS (x, y-);

     else    DFS (x, y+);

     printf (" %d", y);

 }

 int main(void)        //URAL 1029 Ministry

 {

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

     int n, m;

     while (scanf ("%d%d", &n, &m) == )

     {

         memset (pre, -, sizeof (pre));

         memset (dp, , sizeof (dp));

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

             for (int j=; j<=m; ++j)    scanf ("%d", &a[i][j]);

         for (int i=; i<=m; ++i)    {dp[][i] = a[][i];    pre[][i] = -;}

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

         {

             for (int j=; j<=m; ++j)    {dp[i][j] = dp[i-][j] + a[i][j];    pre[i][j] = ;}

             for (int j=; j<=m; ++j)

             {

                 if (dp[i][j] > dp[i][j-] + a[i][j])    {dp[i][j] = dp[i][j-] + a[i][j];    pre[i][j] = ;}

             }

             for (int j=m-; j>=; --j)

             {

                 if (dp[i][j] > dp[i][j+] + a[i][j])    {dp[i][j] = dp[i][j+] + a[i][j];    pre[i][j] = ;}

             }

         }

         int mn = INF;    int x = n;    int y = ;

         for (int i=; i<=m; ++i)

         {

             if (mn > dp[n][i])    {mn = dp[n][i];    y = i;}

         }

         //printf ("%d\n", mn);

         DFS (x, y);    puts ("");

     }

     return ;

 }