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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
inline void write(int x)
{
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
inline void pr1(int x){write(x),putchar(' ');}
inline void pr2(int x){write(x),puts("");}
int a[510][510];
long long f[510][510],sum[510][510],cnt[510][510];
inline long long mymin(long long a,long long b)
{
	if(a>b)swap(a,b);
	if(a==0)return b;
	else return a;
}
int main()
{
    //freopen("a.in","r",stdin);
    //freopen("a.out","w",stdout);
    int n=read();
    for(int i=1;i<=n;i++)
    {
    	for(int j=1;j<=n;j++)if(i!=j)a[i][j]=read();
    }
    for(int i=1;i<=n;i++)
    {
    	for(int j=1;j<=n;j++)sum[i][j]=sum[i][j-1]+a[i][j];
    }
    for(int i=1;i<=n;i++)
    {
    	for(int j=1;j<=n;j++)cnt[i][j]=cnt[i][j-1]+sum[j][i]-sum[j][j-1];
    }
    for(int i=n-1;i>=1;i--)
    {
    	for(int j=i+1;j<=n;j++)
    	{
    		f[i][j]=cnt[j-1][j-1]-cnt[j-1][i-1];
    		long long tot=0,mn=0,uf=0;
    		for(int k=n;k>=j;k--)
    		{
    			tot+=sum[k][j-1]-sum[k][i-1];
    			uf+=sum[k][n]-sum[k][k-1];
    			if(k==j)mn=mymin(mn,uf);
    			else mn=mymin(mn,f[j][k]+tot);
    		}f[i][j]+=mn;
    	}
    }long long ans=f[1][2];
    for(int i=3;i<=n;i++)ans=min(ans,f[1][i]);
    printf("%lld\n",ans);
	return 0;
}