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>
#define debug(x) cerr << #x << " " << (x) << endl
using namespace std;

const int N = 505;
const long long inf = 1000000000000000ll;

int n;
long long a[N][N], pre1[N][N], pre2[N][N], f[N][N], ans = 0ll, mx = -inf;

long long query1 (int l1, int r1, int l2, int r2) {
	return pre1[r1][r2] - pre1[l1 - 1][r2] - pre1[r1][l2 - 1] + pre1[l1 - 1][l2 - 1];
}

long long query2 (int l1, int r1, int l2, int r2) {
	return pre2[r1][r2] - pre2[l1 - 1][r2] - pre2[r1][l2 - 1] + pre2[l1 - 1][l2 - 1];
}

int main () {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		a[i][i] = 0ll;
		for (int j = 1; j < n; j++) {
			int u = i, v = j;
			if (v >= i) v++;
			scanf("%lld", &a[u][v]), ans += a[u][v];
		}
	}

	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= n; j++) pre1[i][j] = pre2[i][j] = 0ll;
	}

	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			long long val1 = i > j ? max(a[i][j], 0ll) : 0ll;
			long long val2 = i < j ? max(a[i][j], 0ll) : 0ll;
			pre1[i][j] = pre1[i - 1][j] + pre1[i][j - 1] - pre1[i - 1][j - 1] + val1;
			pre2[i][j] = pre2[i - 1][j] + pre2[i][j - 1] - pre2[i - 1][j - 1] + val2;
		}
	}

	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= i; j++) {
			f[i][j] = -inf;
			if (j <= 1) f[i][j] = max(f[i][j], query1(1, i, 1, i));
			else {
				for (int k = 1; k < j; k++) f[i][j] = max(f[i][j], f[j - 1][k] + query1(j, i, k, i) + query2(1, j - 1, j, i));
			}
			if (i == n) mx = max(mx, f[n][j]);
		}
	}

	ans -= mx;
	printf("%lld\n", ans);
	return 0;
}