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>
#include<algorithm>
using namespace std;
typedef long long ll;
const int MAXN = 500;
const ll INF = ll(1E18);
int A[MAXN + 5][MAXN + 5], B[MAXN + 5][MAXN + 5], n;
// A[i][j] -> (p[i] + p[i+1] + ... p[j] >= 1) (i <= j)
// B[i][j] -> (p[i] + p[i+1] + ... p[j] <= 1) (i <= j)
ll SA[MAXN + 5][MAXN + 5], SB[MAXN + 5][MAXN + 5];
// SA[i][j] -> A[1][j] + A[2][j] + ... + A[i][j]
// SB[i][j] -> B[1][j] + B[2][j] + ... + B[i][j]
ll dp[2][MAXN + 5][MAXN + 5], f[2][MAXN + 5], g[2];
int main() {
	scanf("%d", &n);
	for(int i=1;i<=n;i++) {
		for(int j=1;j<=i-1;j++) scanf("%d", &B[j][i-1]);
		for(int j=i+1;j<=n;j++) scanf("%d", &A[i][j-1]);
	}
	for(int j=1;j<=n;j++)
		for(int i=1;i<=n;i++)
			SA[i][j] = A[i][j] + SA[i-1][j], SB[i][j] = B[i][j] + SB[i-1][j];
	n--;
	for(int j=1;j<=n;j++) {
		for(int k=j+1;k<=n;k++)
			dp[0][j][k] = INF;
		f[0][j] = INF;
	}
	g[0] = 0;
	for(int i=1;i<=n;i++) {
		for(int j=1;j<i;j++) {
			for(int k=j+1;k<i;k++)
				dp[1][j][k] = dp[0][j][k], dp[0][j][k] = INF;
			f[1][j] = f[0][j], f[0][j] = INF;
		}
		g[1] = g[0], g[0] = INF;
		for(int j=1;j<i;j++) {
			for(int k=j+1;k<i;k++) {
				dp[0][j][k] = min(dp[0][j][k], dp[1][j][k] + SA[i][i] - SA[k][i] + SB[j][i]);
				dp[0][k][i] = min(dp[0][k][i], dp[1][j][k] + SB[k][i]);
			}
			f[0][j] = min(f[0][j], f[1][j] + SA[i][i] - SA[j][i]);
			dp[0][j][i] = min(dp[0][j][i], f[1][j] + SB[j][i]);
		}
		g[0] = min(g[0], g[1] + SA[i][i]);
		f[0][i] = min(f[0][i], g[1]);
	}
	ll mn = INF;
	for(int j=1;j<=n;j++) {
		for(int k=j+1;k<=n;k++)
			mn = min(mn, dp[0][j][k]);
		mn = min(mn, f[0][j]);
	}
	mn = min(mn, g[0]);
	printf("%lld\n", mn);
}