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>
#define ll long long
#define ld long double
#define db double
#define pint pair<int,int>
#define mk(x,y) make_pair(x,y)
#define fir first
#define sec second
#define Rep(x,y,z) for(int x=y;x<=z;x++)
#define Red(x,y,z) for(int x=y;x>=z;x--)
using namespace std;
const int MAXN=505;
char buf[1<<12],*pp1=buf,*pp2=buf,nc;int ny;
inline char gc() {return pp1==pp2&&(pp2=(pp1=buf)+fread(buf,1,1<<12,stdin),pp1==pp2)?EOF:*pp1++;}
//inline char gc(){return getchar();}
inline int read(){
	int x=0;for(ny=1;nc=gc(),(nc<48||nc>57)&&nc!=EOF;)if(nc==45)ny=-1;if(nc<0)return nc;
	for(x=nc-48;nc=gc(),47<nc&&nc<58&&nc!=EOF;x=(x<<3)+(x<<1)+(nc^48));return x*ny;
}
int a[MAXN][MAXN],n;ll Sum[2][MAXN][MAXN],F[MAXN][MAXN];
inline ll Get(int opt,int lx,int rx,int ly,int ry){return Sum[opt][rx][ry]-Sum[opt][lx-1][ry]-Sum[opt][rx][ly-1]+Sum[opt][lx-1][ly-1];}
int main(){
//	freopen("std.in","r",stdin);
//	freopen("std.out","w",stdout);
	n=read();
	Rep(i,1,n)Rep(j,1,n)if(j!=i)a[i][j]=read();
	Rep(i,1,n)Rep(j,1,n){
		if(i<j)Sum[0][i][j]=a[i][j];if(i>j)Sum[1][i][j]=a[i][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];
	}memset(F,0x3f,sizeof(F)),F[0][0]=0;
	Rep(i,0,n)Rep(j,0,max(i-1,0))Rep(k,i+1,n)F[k][i]=min(F[k][i],F[i][j]+Get(1,k+1,n,j+1,i)+Get(0,i+1,k,i+1,k));
	ll ans=1e18;Rep(j,0,n)ans=min(ans,F[n][j]);cout<<ans<<"\n";
	return 0;
}


