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<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<iostream>
#define llong long long
using namespace std;

inline int read()
{
	int x=0; bool f=1; char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=0;
	for(; isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^'0');
	if(f) return x;
	return -x;
}

const int N = 500;
llong a[N+3][N+3];
llong s[2][N+3][N+3];
llong dp[N+3][N+3];
int n;

void update(llong &x,llong y) {x = x<y?x:y;}

llong getsum(int typ,int lx,int rx,int ly,int ry)
{
	return s[typ][rx][ry]-s[typ][lx-1][ry]-s[typ][rx][ly-1]+s[typ][lx-1][ly-1];
}

int main()
{
	scanf("%d",&n);
	for(int i=1; i<=n; i++)
	{
		for(int j=1; j<=n; j++)
		{
			if(j==i) continue;
			scanf("%lld",&a[i][j]);
		}
	}
	for(int i=1; i<=n; i++)
	{
		for(int j=1; j<=n; j++)
		{
			if(i<j) {s[0][i][j] = a[i][j];}
			s[0][i][j] += s[0][i][j-1];
		}
		for(int j=1; j<=n; j++) s[0][i][j] += s[0][i-1][j];
	}
	for(int i=1; i<=n; i++)
	{
		for(int j=1; j<=n; j++)
		{
			if(i>j) {s[1][i][j] = a[i][j];}
			s[1][i][j] += s[1][i][j-1];
		}
		for(int j=1; j<=n; j++) s[1][i][j] += s[1][i-1][j];
	}
	memset(dp,42,sizeof(dp)); 
	dp[0][0] = 0ll;
	for(int i=0; i<=n; i++)
	{
		for(int j=0; j<max(i,1); j++)
		{
//			printf("dp[%d][%d]=%lld\n",i,j,dp[i][j]);
			for(int k=i+1; k<=n; k++)
			{
				llong tmp = dp[i][j]+getsum(1,k+1,n,j+1,i)+getsum(0,i+1,k,i+1,k);
				update(dp[k][i],tmp);
//				printf("update dp[%d][%d]=%lld\n",k,i,tmp);
			}
		}
	}
	llong ans = dp[n][1];
	for(int i=1; i<=n; i++) update(ans,dp[n][i]);
	printf("%lld\n",ans);
	return 0;
}