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
//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define inf 20021225
#define N 510 
using namespace std;
int read()
{
	int s=0,t=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')	t=-1; ch=getchar();}
	while(ch>='0' && ch<='9')	s=s*10+ch-'0',ch=getchar();
	return s*t;
}
ll f[N][N],pre[2][N][N],a[N][N]; int n;
ll calc(int ty,int xl,int xr,int yl,int yr){return pre[ty][xr][yr]-pre[ty][xl-1][yr]-pre[ty][xr][yl-1]+pre[ty][xl-1][yl-1];}
void init()
{
	for(int i=1;i<=n;i++)	for(int j=1;j<=n;j++)
	{
		for(int t=0;t<2;t++)	pre[t][i][j]=pre[t][i-1][j]+pre[t][i][j-1]-pre[t][i-1][j-1];
		pre[i>j][i][j]+=a[i][j];
	}
}
void upd(ll &x,ll y){x=x<y?x:y;}
int main()
{
	n=read();
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<i;j++)	a[i][j]=read();
		for(int j=i+1;j<=n;j++)	a[i][j]=read();
	}
	init(); memset(f,48,sizeof(f)); f[0][0]=0;
	for(int i=0;i<=n;i++)	for(int j=0;j<max(1,i);j++)
		for(int k=i+1;k<=n;k++) if(f[i][j]!=f[0][1])
			upd(f[k][i],f[i][j]+calc(0,i+1,k,i+1,k)+calc(1,k+1,n,j+1,i));
	ll ans=f[0][1];
	for(int i=1;i<=n;i++)	upd(ans,f[n][i]);
	printf("%lld\n",ans);
	return 0;
}