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<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

Int a[540][540];
Int n;
Int dp[540][540];
Int minuss[540][540];
Int pluss[540][540];
Int calc(Int i, Int j, Int k){
    Int res = minuss[j][k] + pluss[0][k] + pluss[i][j] - pluss[0][j] - pluss[i][k];
//    cout << i << " " << j << " " << k << ":" << res << endl;
    return res;
}

int main(){
    cin >> n;
    for(int i = 0;i < n;i++){
        for(int j = 0;j < n;j++){
            if(i == j)continue;
            cin >> a[i][j];
        }
    }
    for(Int k = 2;k <= n;k++){
        for(Int i = 0;i + k <= n;i++){
            Int j = i + k;
            minuss[i][j] = minuss[i][j-1] + minuss[i+1][j] - minuss[i+1][j-1] + a[i][j-1];
            pluss[i][j] = pluss[i][j-1] + pluss[i+1][j] - pluss[i+1][j-1] + a[j-1][i];
//            cout << i << " " << j << " " << minuss[i][j] << " " << pluss[i][j] << endl;
        }
    }
    for(Int i = 1;i <= n;i++){
        dp[0][i] = minuss[0][i];
        
        //      cout << "dp["<< 0 <<"][" << i <<"]: " << dp[0][i] << endl;
    }

    for(Int i = 1;i < n;i++){
        for(Int j = i+1;j <= n;j++){
            dp[i][j] = LINF;
            for(Int k = 0;k < i;k++){
                dp[i][j] = min(dp[i][j], dp[k][i] + calc(k, i, j));
            }
//            cout << "dp["<< i <<"][" << j <<"]: " << dp[i][j] << endl;
        }
    }
    Int res = LINF;
    for(Int i = 0;i < n;i++){
        res = min(res, dp[i][n]);
    }
    cout << res << endl;
    
    return 0;
}