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.

Examples

Input

3
2 1
1 4
3 3


Output

2


Input

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


Output

2


Input

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


Output

2280211
#include <bits/stdc++.h>
using namespace std;

#define LL long long

const int maxN=505;

int n;
LL f[maxN+1][maxN+1],A[maxN+1][maxN+1],s1[maxN+1][maxN+1],s2[maxN+1][maxN+1];

inline LL read()
{
	LL num=0,f=1;
	char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch))num=(num<<3)+(num<<1)+(ch^48),ch=getchar();
	return num*f;
}

inline LL query(int xu,int yu,int xd,int yd,LL sum[maxN+1][maxN+1])
{
	return sum[xd][yd]-sum[xu-1][yd]-sum[xd][yu-1]+sum[xu-1][yu-1];
}

int main()
{

	n=read();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		{
			if(i!=j)A[i][j]=read();
			s1[i][j]=s1[i-1][j]+s1[i][j-1]-s1[i-1][j-1];
			s2[i][j]=s2[i-1][j]+s2[i][j-1]-s2[i-1][j-1];
			if(i<j)s1[i][j]+=A[i][j];
			if(i>j)s2[i][j]+=A[i][j];
		}
	
	for(int i=1;i<=n-1;i++)
	{
		f[i][0]=query(1,1,i+1,i,s2);
		for(int j=0;j<i;j++)
		{
			f[i][i]=max(f[i][i],f[i-1][j]+query(i+1,j+1,i+1,i,s2)+query(1,i+1,i,i+1,s1));
			for(int k=0;k<j;k++)
				f[i][j]=max(f[i][j],f[j-1][k]+query(j+1,k+1,i+1,i,s2)+query(1,j+1,j,i+1,s1));
		}
	}

	LL ans=0;
	for(int i=0;i<=n-1;i++)ans=max(ans,f[n-1][i]);
	printf("%lld",s1[n][n]+s2[n][n]-ans);
}
