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>
int a[510][510];
long long b[510][510], g[510][510];
long long dp[510][510];
inline void upd(long long &v, long long x) {
	if (v > x) v = x;
}
int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			if (i == j) continue;
			scanf("%d", &a[i][j]);
		}
	}
	for (int i = 0; i < n; i++) {
		for (int j = i + 1; j < n; j++) {
			g[i][j] = g[i][j-1];
			for (int k = i; k < j; k++) {
				g[i][j] += a[k][j];
			}
		}
	}
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			b[i+1][j+1] = b[i+1][j] + b[i][j+1] + a[i][j] - b[i][j];
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = i - 1; j > 0; j--) {
			dp[i][j] = 1ll << 60;
			for (int k = j - 1; k >= 0; k--) {
				upd(dp[i][j], dp[j][k] + g[j][i-1] + b[i][k] - b[j][k]);
			}
		}
		dp[i][0] = g[0][i-1];
	}
	long long ans = 1ll << 60;
	for (int i = 0; i < n; i++) {
		upd(ans, dp[n][i]);
	}
	printf("%lld\n", ans);
	return 0;
}
