/*

https://yukicoder.me/problems/no/1028

難読なだけ？
ドローンの移動回数はx座標とy座標の差の最大値
凸関数っぽいので三分探索できる

計算量はO(N*N*logN)

*/

#include <bits/stdc++.h>
using namespace std;

#define rep(i,n,m) for(int (i)=(n);(i)<(m);(i)++)
#define rrep(i,n,m) for(int (i)=(n);(i)>(m);(i)--)
using ll = long long;
const ll mod = 998244353;

int main(){

    int N;
    cin >> N;

    vector<vector<pair<int,int>>> a(N,vector<pair<int,int>> (0));

    int na;
    rep(i,0,N){
        rep(j,0,N){
            cin >> na;
            a[na-1].push_back(make_pair(i,j));
        }
    }

    int ans = 0,l,r,m1,m2,s1,s2;

    rep(i,0,N){
        
        l = 0;
        r = N+1;

        while (r-l > 3){
            
            m1 = (l*2+r)/3;
            m2 = (l+r*2)/3;
            s1 = 0;
            s2 = 0;

            rep (j,0,N){
                s1 += max(abs(m1-a[i][j].first) , a[i][j].second);
            }

            rep (j,0,N){
                s2 += max(abs(m2-a[i][j].first) , a[i][j].second);
            }

            if (s1 > s2){
                l = m1;
            }else{
                r = m2;
            }
        }

        int rmin = 1000000000;
        rep(k,l-2,r+2){
            s1 = 0;
            rep (j,0,N){
                s1 += max(abs(k-a[i][j].first) , a[i][j].second);
            }
            rmin = min(rmin,s1);
        }

        ans += rmin;
    }

    cout << ans << endl;
}