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<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
int n;
#define Maxn 505
ll val1[Maxn][Maxn];//below
ll val2[Maxn][Maxn];//up
ll f[Maxn][Maxn];
ll sum[Maxn];
inline int calc(int x){return x*(x+1)/2;}
inline void rd(int &x){
	x=0;char ch=getchar();
	while(ch<'0'||ch>'9')ch=getchar();
	while(ch>='0'&&ch<='9'){
		x=x*10+ch-'0';
		ch=getchar();
	}
}
int main(){
	ll All=0;
	rd(n);
	int x;
	for(register int i=1;i<=n;++i)
	    for(register int j=1;j<=n;++j)
	       if(j!=i){
	       	    rd(x);
	       	    All+=x;
	       	    if(j<i)val1[j][i-1]=x;
	       	    else val2[i][j-1]=x;
		   }
	n--;
	for(register int i=1;i<=n;++i)
	    for(register int j=2;j<=i;++j){
	    	val1[j][i]+=val1[j-1][i];
	    	val2[j][i]+=val2[j-1][i];
		}
	ll Ans=0;
	ll Ha=0;
	for(register int i=1;i<=n;++i){
		Ha+=val1[i][i];
	    f[i][0]=Ha+val2[i][i];
		for(register int j=0;j<i;++j){
			ll tmp=0;
		    for(register int k=i+1;k<=n;++k){
		    	f[k][i]=max(f[k][i],f[i][j]+tmp+val1[k][k]-val1[i][k]+val2[k][k]);
                tmp+=val1[k][k]-val1[j][k]+val2[i][k];
			}
		}
	}
	for(register int i=1;i<=n;++i)sum[i]=val1[i][i]+sum[i-1];
	for(register int i=1;i<=n;++i)
	    for(register int j=i+1;j<=n;++j){
	    	val1[i][j]+=val1[i][j-1];
	    	val2[i][j]+=val2[i][j-1];
		}
	for(register int i=1;i<=n;++i)
	    for(register int j=0;j<i;++j)
	    	Ans=max(Ans,f[i][j]-val1[j][n]+val1[j][i]+val2[i][n]-val2[i][i]+sum[n]-sum[i]);
    printf("%lld\n",All-Ans);
	return 0;	    
}