Problem:
Score : 1200 points

Problem Statement

We have a weighted directed graph with N vertices numbered 0 to N-1.

The graph initially has N-1 edges. The i-th edge (0 \leq i \leq N-2) is directed from Vertex i to Vertex i+1 and has a weight of 0.

Snuke will now add a new edge (i → j) for every pair i, j (0 \leq i,j \leq N-1,\ i \neq j).
The weight of the edge will be -1 if i < j, and 1 otherwise.

Ringo is a boy. A negative cycle (a cycle whose total weight is negative) in a graph makes him sad.
He will delete some of the edges added by Snuke so that the graph will no longer contain a negative cycle.
The cost of deleting the edge (i → j) is A_{i,j}. He cannot delete edges that have been present from the beginning.

Find the minimum total cost required to achieve Ringo's objective.

Constraints

3 \leq N \leq 500

1 \leq A_{i,j} \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_{0,1} A_{0,2} A_{0,3} \cdots A_{0,N-1}
A_{1,0} A_{1,2} A_{1,3} \cdots A_{1,N-1}
A_{2,0} A_{2,1} A_{2,3} \cdots A_{2,N-1}
\vdots
A_{N-1,0} A_{N-1,1} A_{N-1,2} \cdots A_{N-1,N-2}

Output

Print the minimum total cost required to achieve Ringo's objective.

Sample Input 1

3
2 1
1 4
3 3

Sample Output 1

2

If we delete the edge (0 → 1) added by Snuke, the graph will no longer contain a negative cycle.
The cost will be 2 in this case, which is the minimum possible.

Sample Input 2

4
1 1 1
1 1 1
1 1 1
1 1 1

Sample Output 2

2

If we delete the edges (1 → 2) and (3 → 0) added by Snuke, the graph will no longer contain a negative cycle.
The cost will be 1+1=2 in this case, which is the minimum possible.

Sample Input 3

10
190587 2038070 142162180 88207341 215145790 38 2 5 20
32047998 21426 4177178 52 734621629 2596 102224223 5 1864
41 481241221 1518272 51 772 146 8805349 3243297 449
918151 126080576 5186563 46354 6646 491776 5750138 2897 161
3656 7551068 2919714 43035419 495 3408 26 3317 2698
455357 3 12 1857 5459 7870 4123856 2402 258
3 25700 16191 102120 971821039 52375 40449 20548149 16186673
2 16 130300357 18 6574485 29175 179 1693 2681
99 833 131 2 414045824 57357 56 302669472 95
8408 7 1266941 60620177 129747 41382505 38966 187 5151064

Sample Output 3

2280211

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 998244353;
const ll INF = mod * mod;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef pair<ll, ll> LP;
typedef vector<LP> vec;
typedef vector<string> svec;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-8;

ll sum[501][501];
ll dp[502][502];
ll cop[502][502];


void solve() {
	int n; cin >> n;
	vector<vector<ll>> a(n);
	rep(i, n) {
		a[i].resize(n);
		rep(j, n) {
			if (i == j)a[i][j] = 0;
			else cin >> a[i][j];
		}
	}

	//calc sum
	rep(i, n) {
		ll s = 0;
		for (int j = i + 1; j < n; j++) {
			for (int k = i; k < j; k++) {
				s += a[k][j];
			}
			sum[i][j] = s;
		}
	}

	rep(i, n + 1)rep(j, n + 1)dp[i][j] = INF;
	dp[0][0] = 0;
	{
		ll s = 0;
		for (int j = n-1; j>0; j--) {
			for (int k = 0; k <= j; k++) {
				dp[j][k] = s + sum[0][k] + sum[k + 1][j];
			}

			s += a[j][0];
		}
		/*cout << "0 is result " << endl;
		rep(j, n)rep(k, n) {
			cout << j << " " << k << " " << dp[j][k] << endl;
		}*/
	}
	for (int i = 1; i < n-1; i++) {
		rep(j, n + 1)rep(k, n + 1)cop[j][k] = INF;
		for (int j = n - 1; j >= i; j--) {
			for (int k = i-1; k <= j; k++) {
				if (dp[j][k] == INF)continue;
				ll s = 0;
				if (k >= i) {
					for (int l = n - 1; l >= j; l--) {
						cop[l][k] = min(cop[l][k], s+dp[j][k] + sum[k + 1][l] - sum[k + 1][j]);
						s += a[l][i];
					}
				}
				else {
					for (int l = n - 1; l >= j; l--) {
						for (int m = j; m <= l; m++) {
							cop[l][m] = min(cop[l][m], s + dp[j][k] + sum[i][m] - sum[i][j] + sum[m+1][l]);
						}
						
						s += a[l][i];
					}
				}
			}
		}
		rep(j, n + 1)rep(k, n + 1)dp[j][k]=cop[j][k];

		/*cout << i << " is result " << endl;
		rep(j, n)rep(k, n) {
			cout << j << " " << k << " " << dp[j][k] << endl;
		}*/
	}
	ll ans = INF;
	/*rep(i, n)rep(j, n) {
		cout << i << " " << j << " " << dp[i][j] << endl;
	}*/
	rep(i, n+1)rep(j, n+1)ans = min(ans, dp[i][j]);
	cout << ans << endl;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(10);
	//init();
	solve();
	//cout << "finish" << endl;
	//stop
	return 0;
}