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 rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;

const int N=505;
ll a[N][N],b[N][N],c[N][N],f[N][N],bas,mx=-1e18,sum;
int n;

void cmax(ll &x,ll y){
	if(y>x)x=y;
}

int main(){
	scanf("%d",&n);
	rep(i,1,n)rep(j,1,n)	
		if(i!=j){
			scanf("%lld",&a[i][j]);
			if(i<j)b[i][j]=a[i][j],bas+=a[i][j];
			else c[j][i]=a[i][j];
			sum+=a[i][j];
		}
	per(i,n,1)rep(j,i+2,n){
		b[i][j]+=b[i][j-1]+b[i+1][j]-b[i+1][j-1];
		c[i][j]+=c[i][j-1]+c[i+1][j]-c[i+1][j-1];
	}
	memset(f,0xc0,sizeof f);
	rep(i,1,n)f[1][i]=c[1][i]-b[1][i];
	rep(i,1,n)rep(j,i,n)rep(k,j+1,n)
		cmax(f[j+1][k],f[i][j]+c[i][k]-c[i][j]-b[j+1][k]);
	rep(i,1,n)cmax(mx,f[i][n]);
	cout<<sum-(mx+bas)<<endl;
	return 0;
}