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 <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <fstream>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

typedef long long ll;

const int MAXN = 505;

int n;
int a[MAXN][MAXN];

ll ans;
ll in[MAXN][MAXN];
ll out[MAXN][MAXN];
ll f[2][MAXN][MAXN];

int main()
{
	scanf("%d",&n);
	for (int i = 1;i <= n;i++)
		for (int j = 1;j <= n;j++)
			if (i != j)
				scanf("%d",&a[i][j]);
	for (int i = 1;i <= n;i++)
		for (int j = 1;j < i;j++)
		{
			in[i][j] = in[i][j - 1] + a[j][i];
			out[i][j] = out[i][j - 1] + a[i][j];
		}
	for (int i = 0;i < 2;i++)
		for (int j = 0;j <= i;j++)
			for (int k = 0;k <= j;k++)
				f[i][j][k] = 1e18;
	f[0][0][0] = 0;
	for (int i = 0;i < n - 1;i++)
	{
		for (int j = 0;j <= i + 1;j++)
			for (int k = 0;k <= j;k++)
				f[~i & 1][j][k] = 1e18;
		for (int j = 0;j <= i;j++)
			for (int k = 0;k <= j;k++)
				if (f[i & 1][j][k] != 1e18)
				{
					f[~i & 1][j][k] = min(f[~i & 1][j][k],f[i & 1][j][k] + in[i + 2][i + 1] - in[i + 2][j] + out[i + 2][k]);
					f[~i & 1][i + 1][j] = min(f[~i & 1][i + 1][j],f[i & 1][j][k] + out[i + 2][j]);
				}
	}
	ans = 1e18;
	for (int i = 0;i <= n - 1;i++)
		for (int j = 0;j <= i;j++)
			ans = min(ans,f[~n & 1][i][j]);
	printf("%lld\n",ans);
	return 0;
}