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: Accepted)

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x) {
	x=0; char c=getchar(); int f=1;
	while(!isdigit(c)) { if(c=='-') f=-1; c=getchar(); }
	while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
template <class T> inline void cmin(T &x,T y) { x=min(x,y); }
const int N=510;
ll a[N][N],b[N][N];
ll f[N][N];
int n;
void getsum(ll a[][N]) {
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			a[i][j]+=a[i-1][j]+a[i][j-1]-a[i-1][j-1];
}
ll query(ll a[][N],int x1,int x2,int y1,int y2) {
	if(x2<x1||y2<y1) return 0;
	x1--,y1--;
	return a[x2][y2]-a[x1][y2]-a[x2][y1]+a[x1][y1];
}
int main() {
	rd(n);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j) {
			if(i==j) continue;
			int x; rd(x);
			if(i<j) a[i][j]=x;
			if(i>j) b[i][j]=x;
		}
	getsum(a),getsum(b);
	memset(f,0x3f,sizeof(f));
	for(int i=1;i<=n;++i)
		f[0][i]=query(a,1,i-1,2,i);
	for(int i=0;i<=n-1;++i)
		for(int j=i+1;j<=n-1;++j)
			for(int k=j+1;k<=n-1;++k)
				f[j][k]=min(f[j][k],f[i][j]+query(a,j+1,k-1,j+2,k)+query(b,j+1,k,1,i));
	ll ans=query(a,1,n-1,2,n);
	for(int i=0;i<=n-1;++i)
		for(int j=i+1;j<=n-1;++j) 
			ans=min(ans,f[i][j]+query(a,j+1,n-1,j+2,n)+query(b,j+1,n,1,i));
	printf("%lld",ans);
	return 0;
}