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;
typedef long long LL;
int _max(int x, int y) {return x > y ? x : y;}
LL _min(LL x, LL y) {return x < y ? x : y;}
const int N = 502;
int read() {
	int s = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
	return s * f;
}
void put(LL x) {
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) put(x / 10);
	putchar(x % 10 + '0');
}

LL f[N][N], s[N][N], ss[N][N];
int n, v[N][N];

LL S(int x1, int x2, int y1, int y2) {
	if(x1 > x2 || y1 > y2) return 0;
	return ss[x2][y2] - ss[x2][y1 - 1] - ss[x1 - 1][y2] + ss[x1 - 1][y1 - 1];
}

int main() {
	n = read();
	for(int i = 1; i <= n + 1; i++) {
		for(int j = 1; j <= n + 1; j++) {
			if(i != j && j <= n && i <= n) v[i][j] = read();
			ss[i][j] = v[i][j] + ss[i - 1][j] + ss[i][j - 1] - ss[i - 1][j - 1];
			if(j >= i) s[i][j] = s[i][j - 1] + v[i][j];
		} for(int j = i; j <= n + 1; j++) s[i][j] += s[i - 1][j];
	} memset(f, -1, sizeof(f)); f[0][0] = 0;
	for(int i = 0; i <= n; i++) for(int j = i; j <= n; j++) if(f[i][j] != -1){
		for(int k = j + 1; k <= n + 1; k++) {
			if(f[j][k] == -1) f[j][k] = f[i][j] + S(k + 1, n, i + 1, j) + s[k][k] - s[j][k];
			else f[j][k] = _min(f[j][k], f[i][j] + S(k + 1, n, i + 1, j) + s[k][k] - s[j][k]);
		}
	} LL ans = f[0][n + 1];
	for(int i = 1; i <= n; i++) ans = _min(ans, f[i][n + 1]);
	put(ans), puts("");
	return 0;
}

