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 <algorithm>
// #include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#include <bitset>
using namespace std;
typedef long long LL;
typedef double db;
const int oo=2139063143;
const int N=510;
const int P=1000000007;
const db eps=1e-7;
#define pritnf printf
//char buf[1<<22],*p1=buf,*p2=buf,obuf[1<<22],*O=obuf;
//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
template <typename T>
inline void sc (T &x)
{
    x=0; static int p; p=1; static char c; c=getchar();
    while (!isdigit(c)) { if (c=='-') p=-1; c=getchar(); }
    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }
    x*=p;
}
template <typename T>
inline void print (T x)
{
    if (x< 0) putchar('-'),x=-x;
    if (x>=10) print(x/10);
    putchar(x%10+'0');
}
template <typename T>
inline void pr (T x) { print(x),putchar('\n'); }
template <typename T1,typename T2>
void chkmax (T1 &A,T2 B) { A=A>=B?A:B; }
template <typename T1,typename T2>
void chkmin (T1 &A,T2 B) { A=A<=B?A:B; }
int n; LL a[N][N],b[N][N];
LL get (int k,int j,int i) { return b[j+1][i]+a[n][j]-a[n][k]-a[i][j]+a[i][k]; }
LL f[N][N];
int main ()
{
    // freopen (".in","r",stdin);
    // freopen (".out","w",stdout);
	sc(n);
	for (int i=1; i<=n; i++)
		for (int j=1; j<=n; j++)
			if (i!=j) sc(a[i][j]);
	for (int j=1; j<=n; j++)
		for (int i=j; i>=1; i--)
			b[i][j]=a[i][j]+b[i+1][j]+b[i][j-1]-b[i+1][j-1];
	for (int i=1; i<=n; i++)
		for (int j=1; j<=n; j++)
			a[i][j]+=a[i][j-1]+a[i-1][j]-a[i-1][j-1];
	LL ans=1e18;
	for (int i=1; i< n; i++)
	{
		f[i][0]=get (0,0,i);
		chkmin (ans,f[i][0]+get (0,i,n));
		for (int j=1; j< i; j++)
		{
			f[i][j]=1e18;
			for (int k=0; k< j; k++)
				chkmin (f[i][j],f[j][k]+get (k,j,i));
			chkmin (ans,f[i][j]+get (j,i,n));
		}
	}
	pr(ans);

    return 0;
}
