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>
#pragma GCC optimize("O3")
using namespace std;
#ifdef _debug
#define dout(i) cout << #i << ' ' << i << ' '
#else 
#define dout(i) //
#endif
using ll = long long;
using ull = unsigned long long;
using ul = unsigned;
using db = double;
int n;
const int maxn = 600;
ll dp[maxn][maxn];
ll w[maxn][maxn];
ll p[maxn][maxn];
ll to[maxn][maxn];
const ll inf = 1ll << 59;
inline ll get(int a, int b, int c, int d){
	return p[b][d] - (c == 0 ? 0 : p[b][c-1]) - (a == 0 ? 0 : p[a-1][d]) + (a == 0 || c == 0 ? 0 : p[a-1][c-1]);
}
inline ll get_to(int l, int r){
	return to[r][r] - to[l-1][r] - to[r][l-1] + to[l-1][l-1];
} 
int main(){
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> n;
	for(int i = 1;i <= n;++i)for(int j = 1;j <= n;++j)if(i != j)cin >> w[i][j];
	for(int i = 1;i <= n;++i)for(int j = 1;j <= n;++j)p[i][j] = p[i-1][j] + p[i][j-1] - p[i-1][j-1] + w[i][j];
	for(int i = 1;i <= n;++i)for(int j = 1;j <= n;++j)to[i][j] = to[i-1][j] + to[i][j-1] - to[i-1][j-1] + (i < j ? w[i][j] : 0);
	for(int i = 1;i <= n;++i){
		dp[i][1] = get_to(1, i);
		for(int j = 2;j <= i;++j){
			dp[i][j] = inf;
			for(int k = 1;k < j;++k){
				dp[i][j] = min(dp[i][j], dp[j-1][k] + get_to(j, i) + get(j, i, 0, k-1));
			}
		}
	}
	cout << *min_element(dp[n]+1, dp[n]+n+1) << '\n';
}