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<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define ll long long
#define inf 0x3f3f3f3f
#define maxn 510
inline ll read()
{
	ll x=0; char c=getchar(),f=1;
	for(;c<'0'||'9'<c;c=getchar())if(c=='-')f=-1;
	for(;'0'<=c&&c<='9';c=getchar())x=x*10+c-'0';
	return x*f;
}
inline void write(ll x)
{
	static char buf[20];
	int len=0;
	if(x<0)putchar('-'),x=-x;
	for(;x;x/=10)buf[len++]=x%10+'0';
	if(!len)putchar('0');
	else while(len)putchar(buf[--len]);
}
inline void writesp(ll x){write(x); putchar(' ');}
inline void writeln(ll x){write(x); putchar('\n');}
int a[maxn][maxn];
ll sum1[maxn][maxn],sum2[maxn][maxn],f[maxn][maxn];
int n;
ll query1(int l1,int r1,int l2,int r2){return sum1[r1][r2]-sum1[r1][l2-1]-sum1[l1-1][r2]+sum1[l1-1][l2-1];}
ll query2(int l1,int r1,int l2,int r2){return sum2[r1][r2]-sum2[r1][l2-1]-sum2[l1-1][r2]+sum2[l1-1][l2-1];}
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();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++){
			sum1[i][j]=sum1[i][j-1]+sum1[i-1][j]-sum1[i-1][j-1];
			sum2[i][j]=sum2[i][j-1]+sum2[i-1][j]-sum2[i-1][j-1];
			if(i<j)sum1[i][j]+=a[i][j];
			else sum2[i][j]+=a[i][j];
		}
	memset(f,0x3f,sizeof(f));
	for(int i=1;i<=n;i++)
		f[i][1]=query1(1,i,1,i);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=i;j++)
			for(int k=1;k<j;k++)
				f[i][j]=std::min(f[i][j],f[j-1][k]+query1(j,i,j,i)+query2(i+1,n,k,j-1));
	ll ans=(1ll<<60);
	for(int i=1;i<=n;i++)
		ans=std::min(ans,f[n][i]);
	writeln(ans);
	// for(int i=1;i<=n;i++){
	// 	for(int j=1;j<=n;j++)
	// 		writesp(f[i][j]);
	// 	putchar('\n');
	// }
	return 0;
}