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<bits/stdc++.h>
using namespace std;
int v[510][510],n;
long long f[510][510],sp[510][510],ss[510][510],ans=1e18;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)for(int j=1;j<=n;++j)if(i^j)scanf("%d",v[i]+j);
	for(int i=1;i<=n;++i)for(int j=n;j>i;--j)
		ss[i][j]=ss[i-1][j]+ss[i][j+1]-ss[i-1][j+1]+v[j][i];
	for(int i=n;i;--i)for(int j=i+1;j<=n;++j)
		sp[i][j]=sp[i+1][j]+sp[i][j-1]-sp[i+1][j-1]+v[i][j];
	for(int i=0;i<=n;++i)for(int j=0;j<=n;++j)f[i][j]=1e18;
	for(int i=1;i<=n;++i)f[i][0]=sp[1][i];
	for(int i=1;i<n;++i)for(int j=0;j<i;++j)for(int k=i+1;k<=n;++k)
		f[k][i]=min(f[k][i],f[i][j]+ss[i][k+1]-ss[j][k+1]+sp[i+1][k]);
	for(int i=0;i<n;++i)ans=min(ans,f[n][i]);
	printf("%lld",ans);
}