#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 21, inf = 0x3f3f3f3f;
int n, m, a[N][N], dp[1 << N][N];

/*
状态：dp[i][j] 表示已经经过i集合的点，且最后在j点的最短 Hamilton 路径的长度.
初始化：dp[i][j]=inf, dp[i][i]=0
转移：k->j, dp[i][j] = min(dp[i][j], dp[i^(1<<j)][k]+a[k][j]);
目标：dp[(1<<n)-1][n-1]
*/
int main() {
    cin >> n;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++) cin >> a[i][j];
    memset(dp, 0x3f, sizeof dp);
    dp[1][0] = 0;
    for (int i = 1; i < (1 << n); i++)
        for (int j = 0; j < n; j++) if (i >> j & 1)
            for (int k = 0; k < n; k++) if (i ^ (1 << j) >> k & 1)
                dp[i][j] = min(dp[i][j], dp[i ^ (1 << j)][k] + a[k][j]);
    cout << dp[(1 << n) - 1][n - 1] << endl;
}