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<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<ctime>
#include<map>
#include<bitset>
#include<set>
#define LL long long
#define mp(x,y) make_pair(x,y)
#define pll pair<long long,long long>
#define pii pair<int,int>
using namespace std;
inline LL read()
{
	LL f=1,x=0;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;
}
int stack[20];
template<typename T>inline void write(T x)
{
	if(x<0){putchar('-');x=-x;}
    if(!x){putchar('0');return;}
    int top=0;
    while(x)stack[++top]=x%10,x/=10;
    while(top)putchar(stack[top--]+'0');
}
template<typename T>inline void pr1(T x){write(x);putchar(' ');}
template<typename T>inline void pr2(T x){write(x);putchar('\n');}
const int MAXN=505;
int a[MAXN][MAXN],n;
LL f[MAXN][MAXN],g[MAXN],s1[MAXN][MAXN],s2[MAXN][MAXN];
int s[MAXN];
int main()
{
//	freopen("d.in","r",stdin);
//	freopen("d.out","w",stdout);
	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++)s1[i][j]=s1[i][j-1]+a[i][j];
	for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)s2[j][i]=s2[j][i-1]+a[i][j];
	memset(f,63,sizeof(f));LL INF=f[0][0];
	for(int i=1;i<n;i++)
	{
		f[i][0]=0;
		for(int j=1;j<=i;j++)f[i][0]+=s1[j][i]-s1[j][j];
	}
	for(int i=1;i<n;i++)for(int j=0;j<i;j++)if(f[i][j]!=INF)
	{
		LL sum=0;
		for(int k=i+1;k<=n;k++)
		{
			sum+=s2[k][k]-s2[k][i]+s1[k][j];
			f[k][i]=min(f[k][i],f[i][j]+sum);
		}
	}
	LL ans=(1LL<<63-1);
	for(int i=0;i<n;i++)ans=min(ans,f[n][i]);
	pr2(ans);
	return 0;
}
