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<algorithm>
#include<vector>
#include<set>
#include<stack>
#include<cmath>
#include<cstring>
using namespace std;
#define MAXN 500
#define LL long long
#define INF 1000000000000000000
int n;
LL sum[2][MAXN+5][MAXN+5],dp[MAXN+5][MAXN+5];
LL read(){
   	LL x=0,F=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')F=-1;c=getchar();}
   	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*F;
}
LL get_sum(int x1,int x2,int y1,int y2,int id){
	return sum[id][x2][y2]-sum[id][x1-1][y2]-sum[id][x2][y1-1]+sum[id][x1-1][y1-1];
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		if(i!=j)sum[(i>j)][i][j]=read();
	for(int i=0;i<=n;i++)
		for(int j=0;j<=n;j++)dp[i][j]=INF;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++){
			sum[0][i][j]+=sum[0][i-1][j]+sum[0][i][j-1]-sum[0][i-1][j-1];
			sum[1][i][j]+=sum[1][i-1][j]+sum[1][i][j-1]-sum[1][i-1][j-1];
		}
	dp[0][0]=0; 
	for(int i=0;i<=n;i++)
		for(int pre=0;pre<max(1,i);pre++)
			for(int nxt=i+1;nxt<=n;nxt++)
			dp[nxt][i]=min(dp[nxt][i],dp[i][pre]+get_sum(nxt+1,n,pre+1,i,1)+get_sum(i+1,nxt,i+1,nxt,0));
	LL ans=dp[n][1];
	for(int i=1;i<=n;i++)ans=min(ans,dp[n][i]);
	printf("%lld",ans);
}