Problem:
Score : 700 points

Problem Statement

Tonight, in your favourite cinema they are giving the movie Joker and all seats are occupied. In the cinema there are N rows with N seats each, forming an N\times N square. We denote with 1, 2,\dots, N the viewers in the first row (from left to right); with N+1, \dots, 2N the viewers in the second row (from left to right); and so on until the last row, whose viewers are denoted by N^2-N+1,\dots, N^2.

At the end of the movie, the viewers go out of the cinema in a certain order: the i-th viewer leaving her seat is the one denoted by the number P_i. The viewer P_{i+1} waits until viewer P_i has left the cinema before leaving her seat.
To exit from the cinema, a viewer must move from seat to seat until she exits the square of seats (any side of the square is a valid exit). A viewer can move from a seat to one of its 4 adjacent seats (same row or same column).
While leaving the cinema, it might be that a certain viewer x goes through a seat currently occupied by viewer y; in that case viewer y will hate viewer x forever. Each viewer chooses the way that minimizes the number of viewers that will hate her forever.

Compute the number of pairs of viewers (x, y) such that y will hate x forever.

Constraints

2 \le N \le 500

The sequence P_1, P_2, \dots, P_{N^2} is a permutation of \{1, 2, \dots, N^2\}.

Input

The input is given from Standard Input in the format

N
P_1 P_2 \cdots P_{N^2}

Output

If ans is the number of pairs of viewers described in the statement, you should print on Standard Output

ans

Sample Input 1

3
1 3 7 9 5 4 8 6 2

Sample Output 1

1

Before the end of the movie, the viewers are arranged in the cinema as follows:

1 2 3
4 5 6
7 8 9

The first four viewers leaving the cinema (1, 3, 7, 9) can leave the cinema without going through any seat, so they will not be hated by anybody.

Then, viewer 5 must go through one of the seats where viewers 2, 4, 6, 8 are currently seated while leaving the cinema; hence he will be hated by at least one of those viewers.

Finally the remaining viewers can leave the cinema (in the order 4, 8, 6, 2) without going through any occupied seat (actually, they can leave the cinema without going through any seat at all).

Sample Input 2

4
6 7 1 4 13 16 10 9 5 11 12 14 15 2 3 8

Sample Output 2

3

Sample Input 3

6
11 21 35 22 7 36 27 34 8 20 15 13 16 1 24 3 2 17 26 9 18 32 31 23 19 14 4 25 10 29 28 33 12 6 5 30

Sample Output 3

11

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

// clang-format off
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#ifndef bhartiya
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
// #pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#endif

#define int long long
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef gp_hash_table<int, int> umap;
typedef tree<int, null_type, less<int>, rb_tree_tag,
		tree_order_statistics_node_update> oset;
typedef pair<pii, int> piii;
typedef vector<int> vi;
typedef vector<pii> vii;

#define INF 4557430888798830399ll
#define MOD 1000000007
#define EPS 1e-7
#define PI acos(-1)

#define f first
#define s second
#define rsz resize
#define ins insert //
#define ft front() //
#define bk back()
#define pf push_front //
#define pb push_back
#define eb emplace_back //
#define lb lower_bound //
#define ub upper_bound //

#define pb push_back
#define bitc __builtin_popcountll
#define mp make_pair
#define ff first
#define ss second
#define all(ar) ar.begin(), ar.end()
#define sz(x) (int)(x).size()

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)
#define fr(i, a, b) for (int i = (a), _b = (b); i <= _b; i++)
#define rep(i, n) for (int i = 0, _n = (n); i < _n; i++)
#define repr(i, n) for (int i = n - 1; i >= 0; i--)
#define frr(i, a, b) for (int i = (a), _b = (b); i >= _b; i--)
#define fil(ar, val) memset(ar, val, sizeof(ar))  // 0x3f for inf, 0x80 for -INF can also use with pairs

#ifdef bhartiya
template<typename T>
void __p(T a) { cout << a << " "; }
template<typename T>
void __p(std::vector<T> a) { cout << "{ "; for (auto p : a) __p(p); cout << "}"; }
template<typename T, typename F>
void __p(pair<T, F> a) { cout << "{ "; __p(a.first); __p(a.second); cout << "}"; }
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) { __p(a1); __p(a...); }
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
	cout<<name<<" : ";__p(arg1);cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
	int bracket=0,i=0;
	for(; ;i++)
		if(names[i]==','&&bracket==0)
			break;
		else if(names[i]=='(')
			bracket++;
		else if(names[i]==')')
			bracket--;
	const char *comma=names+i;
	cout.write(names,comma-names)<<" : ";
	__p(arg1);
	cout<<"| ";
	__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" "; __f(#__VA_ARGS__, __VA_ARGS__)
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif

mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
inline bool equals(double a, double b) { return fabs(a - b) < 1e-9; }
ll modpow(ll b, ll e, ll mod=MOD) {
    ll ans=1; for(;e;b=b*b%mod,e/=2) if(e&1) ans=ans*b%mod; return ans; }
// clang-format on
typedef pair<int,int> pi;
typedef pair<ll,ll> pl; //

typedef vector<pi> vpi;
// typedef vector<pii> vpl; //

const int maxn = 1000005;

const int SZ = 1<<9;
template<class T> struct node {
	T val = 0; node<T>* c[2];
	node() { c[0] = c[1] = NULL; }
	void upd(int ind, T v, int L = 0, int R = SZ-1) { // add v
		if (L == ind && R == ind) { val += v; return; }
		int M = (L+R)/2;
		if (ind <= M) {
			if (!c[0]) c[0] = new node();
			c[0]->upd(ind,v,L,M);
		} else {
			if (!c[1]) c[1] = new node();
			c[1]->upd(ind,v,M+1,R);
		}
		val = 0; F0R(i,2) if (c[i]) val += c[i]->val;
	}
	T query(int lo, int hi, int L = 0, int R = SZ-1) { // query sum of segment
		if (hi < L || R < lo) return 0;
		if (lo <= L && R <= hi) return val;
		int M = (L+R)/2; T res = 0;
		if (c[0]) res += c[0]->query(lo,hi,L,M);
		if (c[1]) res += c[1]->query(lo,hi,M+1,R);
		return res;
	}
	void UPD(int ind, node* c0, node* c1, int L = 0, int R = SZ-1) {
		if (L != R) {
			int M = (L+R)/2;
			if (ind <= M) {
				if (!c[0]) c[0] = new node();
				c[0]->UPD(ind,c0?c0->c[0]:NULL,c1?c1->c[0]:NULL,L,M);
			} else {
				if (!c[1]) c[1] = new node();
				c[1]->UPD(ind,c0?c0->c[1]:NULL,c1?c1->c[1]:NULL,M+1,R);
			}
		} 
		val = (c0?c0->val:0)+(c1?c1->val:0);
	}
};


template<class T> struct BITseg {
	node<T> seg[SZ];
	BITseg() { F0R(i,SZ) seg[i] = node<T>(); }
	void upd(int x, int y, int v) { // add v
		for (; x < SZ; x += x&-x) seg[x].upd(y,v); }
	T query(int x, int yl, int yr) {
		T res = 0; for (; x; x-=x&-x) res += seg[x].query(yl,yr);
		return res; }
	T query(int xl, int xr, int yl, int yr) { // query sum of rectangle
		return query(xr,yl,yr)-query(xl-1,yl,yr); }
};

void solve() 
{
	BITseg<int> seg;
	int n; cin>>n;
	// trace(n);
	seg.upd(50,50,0);
	for(int i=1; i<=n; i++){
		fr(j,1,n){
			// trace(i,j);
			seg.upd(i,j,1);
		}
	}
	// trace(seg.query(1,1,1,1));
	// trace(n);
	int ans = 0;
	fr(i,1,n*n){
		int lo; cin>>lo;
		int x = (lo-1)/n + 1;
		int y = lo - (x-1)*n;
		// trace(lo, x, y);
		seg.upd(x,y,-1);
		int val = min(seg.query(x,x,1,y), min(seg.query(1,x,y,y), min(seg.query(x,n,y,y), seg.query(x,x,y,n))));
		// trace(val);
		ans += val;
	}
	cout<<ans<<endl;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0), cout.tie(0), cout.precision(10);  // cout<<fixed;
#ifdef bhartiya
    cin.exceptions(cin.failbit);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
#endif
    int t=1;
    // cin>>t; 
    fr(i,1,t){
        // cout<<"Case #"<<i<<": ";
        solve();
    }
    end_routine();
}