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 debug(...) fprintf(stderr,__VA_ARGS__)
#define fi first
#define se second
#define MK make_pair
#define PB push_back
using namespace std;

typedef long long LL;
typedef pair<int ,int> PII;
const int N=510;
const LL oo = 1ll<<60;
LL pre[N][N],pre2[N][N], a[N][N], f[N][N];
template<class T>
void upmin(T&a, T b){if ( b<a) a=b;}

int n, m;
LL Inner(int i, int j){return pre2[j][j]-pre2[i-1][j]-pre2[j][i-1]+pre2[i-1][i-1];}
LL Outer(int i,int j, int k){return pre[k][i-1]-pre[j][i-1];}
int main(){
	#ifdef che
	freopen("input.txt","rt",stdin);
	freopen("output.txt","wt",stdout);
	#endif
	scanf("%d", &n);
	for (int i=1; i<=n; ++i)
		for (int j=1; j<=n; ++j) if ( i!=j){
			scanf("%lld", a[i]+j);
			pre[i][j]= a[i][j];
			if ( i<j) pre2[i][j] = a[i][j];
		}
	for (int i=1; i<=n; ++i)
		for (int j=1; j<=n; ++j) pre[i][j] += pre[i-1][j], pre2[i][j]+=pre2[i-1][j];
	for (int i=1; i<=n; ++i)
		for (int j=1; j<=n; ++j) pre[i][j] += pre[i][j-1], pre2[i][j]+=pre2[i][j-1];
	
	for (int i=1; i<=n; ++i)
		for (int j=i; j<=n; ++j) f[i][j] = oo;
	
	for (int i=1; i<=n; ++i)
		for (int j=i; j<=n; ++j){
			if ( i==1) f[i][j] = Inner(i,j);
			for (int k=j+1; k<=n; ++k) upmin(f[j+1][k], f[i][j] + Inner(j+1,k)+Outer(i,j,k));
			#ifdef che
			debug("f[%d][%d] = %lld\n", i,j,f[i][j]);
			#endif
		}
	LL ans=oo;
	for (int i=1; i<=n; ++i) upmin(ans, f[i][n]);
	printf("%lld\n", ans);
	return 0;
}