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;

#define ll  long long
#define ld  double

#define sz(x)   (int)x.size()
#define all(x)  x.begin(),x.end()

#define pb  emplace_back
#define X   first
#define Y   second

const int   N   = 505;

typedef pair<int,int>   ii;

ll  f[N][N];
ll  s_r[N][N];
ll  s_c[N][N];

int main()  {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    int n;  cin >> n;

    for(int i = 1 ; i <= n ; ++i)
    for(int j = 1 ; j <= n ; ++j)   {
        int x = 0;
        int y = 0;

        if (i < j)  cin >> x;
        if (i > j)  cin >> y;

        s_r[i][j] = x + s_r[i][j - 1] + s_r[i - 1][j] - s_r[i - 1][j - 1];
        s_c[i][j] = y + s_c[i][j - 1] + s_c[i - 1][j] - s_c[i - 1][j - 1];
    }
    for(int i = 0 ; i <= n ; ++i)
    for(int j = 0 ; j <= n ; ++j)
        f[i][j] = 1e18;

    f[0][0] = 0;

    for(int i = 0 ; i <  n ; ++i)
    for(int j = 0 ; j <= i ; ++j)   if (f[i][j] < 1e18)
        for(int k = i + 1 ; k <= n ; ++k)   {
            ll  nxt = f[i][j];

            nxt += s_r[k][k] + s_r[i][i] - s_r[k][i] - s_r[i][k];
            nxt += s_c[n][i] + s_c[k][j] - s_c[k][i] - s_c[n][j];

            if (f[k][i] > nxt)
                f[k][i] = nxt;
        }

    cout << *min_element(f[n],f[n] + 1 + n) << endl;
}