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>
using namespace std;

#define LL long long

const int maxN = 505;

int n;
LL f[maxN + 1][maxN + 1], A[maxN + 1][maxN + 1], s1[maxN + 1][maxN + 1], s2[maxN + 1][maxN + 1];

inline LL read()
{
	LL num = 0, f = 1;
	char ch = getchar();
	while( !isdigit( ch ) ) { if(ch == '-') f = -1; ch = getchar(); }
	while( isdigit( ch ) ) num = (num << 3) + (num << 1) + (ch ^ 48), ch = getchar();
	return num * f;
}

inline LL query(int xu, int yu, int xd, int yd, LL sum[maxN + 1][maxN + 1])
{
	return sum[xd][yd] - sum[xu - 1][yd] - sum[xd][yu - 1] + sum[xu - 1][yu - 1];
}

int main()
{

	n = read();
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
		{
			if(i != j) A[i][j] = read();
			s1[i][j] = s1[i - 1][j] + s1[i][j - 1] - s1[i - 1][j - 1];
			s2[i][j] = s2[i - 1][j] + s2[i][j - 1] - s2[i - 1][j - 1];
			if(i < j) s1[i][j] += A[i][j];
			if(i > j) s2[i][j] += A[i][j];
		}
	
	for(int i = 1; i <= n - 1; i++)
	{
		f[i][0] = query(1, 1, i + 1, i, s2);
		for(int j = 0; j < i; j++)
		{
			f[i][i] = max(f[i][i], f[i - 1][j] + query(i + 1, j + 1, i + 1, i, s2) + query(1, i + 1, i, i + 1, s1));
			for(int k = 0; k < j; k++)
				f[i][j] = max(f[i][j], f[j - 1][k] + query(j + 1, k + 1, i + 1, i, s2) + query(1, j + 1, j, i + 1, s1));
		}
	}

	LL ans = 0;
	for(int i = 0; i <= n - 1; i++) ans = max(ans, f[n - 1][i]);
	printf("%lld", s1[n][n] + s2[n][n] - ans);

	return 0;
}
