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;

namespace TYC
{
	typedef long long ll;

	const int N = 505;
	const ll INF = 0x3f3f3f3f3f3f3f3fLL;

	ll A[N][N], S1[N][N], S2[N][N], F[N][N];

	inline int read()
	{
		int x = 0, f = 0, ch = getchar();
		while (!isdigit(ch))
			f |= (ch == '-'), ch = getchar();
		while (isdigit(ch))
			x = x * 10 + ch - '0', ch = getchar();
		return f ? -x : x;
	}

	inline ll Q(const int x1, const int x2, const int y1, const int y2)
	{
		if (x1 > x2 || y1 > y2)
			return 0;
		else
			return S2[x2][y2] - S2[x1 - 1][y2] - S2[x2][y1 - 1] + S2[x1 - 1][y1 - 1];
	}

	void work()
	{
		int n = read();
		for (int i = 1; i <= n; i++)
			for (int j = 1, v; j <= n; j++)
			{
				v = i == j ? 0 : read();
				A[i][j] = A[i][j - 1] + v;
				S2[i][j] = S2[i - 1][j] + S2[i][j - 1] + v - S2[i - 1][j - 1];
			}
		for (int a = 1; a <= n; a++)
			for (int b = a; b <= n; b++)
				for (int i = a; i <= b; i++)
					S1[a][b] += A[i][b] - A[i][i];
		memset(F, 0x3f, sizeof(F));
		for (int i = 1; i <= n; i++)
			F[i][0] = S1[1][i];
		for (int i = 1; i < n; i++)
			for (int j = 0; j < i; j++)
				if (F[i][j] != INF)
					for (int k = i + 1; k <= n; k++)
						F[k][i] = min(F[k][i], F[i][j] + S1[i + 1][k] + Q(k + 1, n, j + 1, i));
		ll ans = INF;
		for (int i = 0; i <= n; i++)
			ans = min(ans, F[n][i]);
		printf("%lld\n", ans);
	}
}

int main()
{
	TYC::work();
	return 0;
}