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<bits/stdc++.h>
using namespace std;
template<typename T>inline T read(){
	T f=0,x=0;char c=getchar();
	while(!isdigit(c)) f=c=='-',c=getchar();
	while(isdigit(c)) x=x*10+c-48,c=getchar();
	return f?-x:x;
}
#define int long long
namespace run{
	const int N=509;
	int n,A[N][N],f[N][N],sum[N][N],g[N][N];
	int main(){
		n=read<int>();
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++) if(i^j)
				A[i][j]=read<int>();

		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++) sum[i][j]=A[i][j]+sum[i][j-1],g[i][j]=(i>=j?A[i][j]:0)+g[i][j-1];
		for(int j=1;j<=n;j++)
			for(int i=1;i<=n;i++) sum[i][j]+=sum[i-1][j],g[i][j]+=g[i-1][j];
	
		for(int i=1;i<=n;i++) f[0][i]=g[i][i];
		int ans=0;
		for(int i=2;i<=n;i++){
			for(int j=0;j<i;j++)
				for(int k=j+1;k<i;k++){
					f[k][i]=max(f[k][i],f[j][k]+sum[k][i]-sum[k][k]+sum[i][k]-sum[i][j]-sum[k][k]+sum[k][j]+g[i][i]+g[k][k]-g[i][k]-g[k][i]);
					if(i==n) ans=max(ans,f[k][i]);
				}
		}
		printf("%lld\n",sum[n][n]-ans);
		return 0;
	}
}
#undef int
int main(){
#ifdef my
	freopen(".in","r",stdin);
	freopen(".out","w",stdout);
#endif
	return run::main();
}