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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstdlib>
#define LL long long
#define LD long double
using namespace std;
const int NN=500 +117;
const int MM= +117;
int read(){
	int fl=1,x;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if(c=='-'){fl=-1;c=getchar();}
	for(x=0;c>='0'&&c<='9';c=getchar())
		x=(x<<3)+(x<<1)+c-'0';
	return x*fl;
}
void open(){
	freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
}
void close(){
	fclose(stdin);
	fclose(stdout);
}


int m,n;
LL v[2][NN][NN]={};
LL dp[2][NN][NN]={};
void upd(LL &x,LL y){
	x=max(x,y);
}
LL vof(int kd,int l,int r,int x){
	return v[kd][r][x]-v[kd][l-1][x];
}
int main(){
	//open();
	n=read();
	memset(dp,-60,sizeof dp);
	LL sum=0;
	for(int i=1;i<=n;++i){
		for(int j=1;j<=n;++j){
			if(i==j)continue;
			LL x=read();
			sum+=x;
			if(i<j){
				v[1][i][j]=x;
			}
			else v[0][j][i]=x;
		}
	}
	for(int i=1;i<=n;++i){
		for(int j=0;j<=n;++j){
			v[1][i][j]+=v[1][i-1][j];
			v[0][i][j]+=v[0][i-1][j];
		}
	}
	dp[0][0][0]=0;
	int now=0;
	for(int i=1;i<=n;++i){
		now^=1;
		for(int j=0;j<=i;++j){
			for(int l=0;l<=i;++l){
				dp[now][j][l]=0;
			}
		}
		for(int j=0;j<i;++j){
			for(int l=j;l<i;++l){
				upd(dp[now][j][l],dp[now^1][j][l]);
				upd(dp[now][l][i],dp[now^1][j][l]);
			}
		}
		for(int j=0;j<i;++j){
			for(int l=j;l<=i;++l){
				dp[now][j][l]+=vof(0,j+1,i,i+1)+vof(1,1,l,i+1);
			}
		}
	}
	LL ans=0;
	for(int j=0;j<n;++j){
		for(int l=j;l<=n;++l){
			upd(ans,dp[now][j][l]);
		}
	}
	ans=sum-ans;
	printf("%lld\n",ans);
	close();
	return 0;
}