#include<bits/stdc++.h>
using namespace std;
const int N = 105;
int a[N][N], c[N], dp[N];//f[i]第i个点到终点的最小花费
int main(){
    int n, t = 0;
    cin>>n;
    for(int i=0; i<n; i++)
        for(int j=0; j<n; j++)
            cin>>a[i][j];
    for(int i=0; i<n-1; i++) dp[i] = N*N*N;
    dp[n-1] = 0,c[n-1] = 0;
    for(int i=n-2; i>=0; i--){//倒着向前找
        for(int j=i+1; j<n; j++)
            if(a[i][j] && a[i][j] + dp[j] < dp[i]){
                dp[i] = a[i][j] + dp[j];
                c[i] = j;//从i到j的最小花费点
            }
    }
    printf("minlong=%d\n%d", dp[0], 1);
    while(c[t]) {
        printf(" %d", c[t]+1);
        t = c[t];
    }
    return 0;
}