#include <stdio.h>

#include <string.h>

#define M 310

#define inf 0x3f3f3f3f

int n,nx,ny;

int link[M],lx[M],ly[M],slack[M];    //lx,ly为顶标，nx,ny分别为x点集y点集的个数

int visx[M],visy[M],w[M][M];

int DFS(int x)

{

    visx[x] = 1;

    for (int y = 1;y <= ny;y ++)

    {

        if (visy[y])

            continue;

        int t = lx[x] + ly[y] - w[x][y];

        if (t == 0)       //

        {

            visy[y] = 1;

            if (link[y] == -1||DFS(link[y]))

            {

                link[y] = x;

                return 1;

            }

        }

        else if (slack[y] > t)  //不在相等子图中slack 取最小的

            slack[y] = t;

    }

    return 0;

}

int KM()

{

    int i,j;

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

    memset (ly,0,sizeof(ly));

    for (i = 1;i <= nx;i ++)            //lx初始化为与它关联边中最大的

        for (j = 1,lx[i] = -inf;j <= ny;j ++)

            if (w[i][j] > lx[i])

                lx[i] = w[i][j];

    for (int x = 1;x <= nx;x ++)

    {

        for (i = 1;i <= ny;i ++)

            slack[i] = inf;

        while (1)

        {

            memset (visx,0,sizeof(visx));

            memset (visy,0,sizeof(visy));

            if (DFS(x))     //若成功（找到了增广轨），则该点增广完毕。进入下一个点的增广

                break;  //若失败（没有找到增广轨），则须要改变一些点的标号，使得图中可行边的数量添加。

                        //方法为：将全部在增广轨中（就是在增广过程中遍历到）的X方点的标号全部减去一个常数d，

                        //全部在增广轨中的Y方点的标号全部加上一个常数d

            int d = inf;

            for (i = 1;i <= ny;i ++)

                if (!visy[i]&&d > slack[i])

                    d = slack[i];

            for (i = 1;i <= nx;i ++)

                if (visx[i])

                    lx[i] -= d;

            for (i = 1;i <= ny;i ++)  //改动顶标后，要把全部不在交错树中的Y顶点的slack值都减去d

                if (visy[i])

                    ly[i] += d;

                else

                    slack[i] -= d;

        }

    }

    int res = 0;

    for (i = 1;i <= ny;i ++)

        if (link[i] > -1)

            res += w[link[i]][i];

    return res;

}

int main ()

{

    int i,j;

    while (scanf ("%d",&n)!=EOF)

    {

        nx = ny = n;

      //  memset (w,0,sizeof(w));

        for (i = 1;i <= n;i ++)

            for (j = 1;j <= n;j ++)

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

        int ans = KM();

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

    }

    return 0;

}