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>
#define ll long long
using namespace std;
int read()
{
	int x=0;char ch=getchar();
	while (ch<'0'||ch>'9')ch=getchar();
	while (ch>='0'&&ch<='9')x=x*10+(ch^48),ch=getchar();
	return x;
}
const ll inf=1e15;
int n,a[505][505];
ll f[505][505],p[505][505],q[505][505];
ll min(ll a,ll b){return a<b?a:b;}
int main()
{
	n=read();
	for (int i=0;i<n;++i)
	for (int j=0;j<n;++j)
	if (i!=j)
	{
		a[i][j]=read();
		if (i<j)
		f[i][j]=inf;
	}
	for (int i=n-1;i>=0;--i)
	{
		for (int j=i+1;j<=n;++j)
		p[i][j]=p[i][j-1]+a[i][j];
		for (int j=i+1;j<=n;++j)
		p[i][j]+=p[i+1][j];
	}
	for (int i=0;i<n;++i)
	{
		for (int j=n-1;j>i;--j)
		q[i][j]=q[i][j+1]+a[j][i];
		if (i)
		for (int j=n-1;j>i;--j)
		q[i][j]+=q[i-1][j];
	}
	for (int i=1;i<=n;++i)
		for (int j=i+1;j<=n+1;++j)
		f[i][j]=p[0][i-1]+p[i][j-1]+q[i-1][j];
	for (int i=1;i<n;++i)
	for (int j=i+1;j<=n;++j)
	for (int k=j+1;k<=n+1;++k)
	f[j][k]=min(f[j][k],f[i][j]+
	p[j][k-1]+q[j-1][k]-q[i-1][k]);
	printf("%lld\n",f[n][n+1]);
	return 0;
}