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
//
// Created by yamunaku on 2019/08/30.
//

#include <bits/stdc++.h>

using namespace std;

#define rep(i, n) for(int i = 0; i < (n); i++)
#define repl(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perl(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD9 998244353
#define MOD1 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)
#define CST(x) cout<<fixed<<setprecision(x)

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vector<int>> mti;
typedef vector<ll> vl;
typedef vector<vector<ll>> mtl;

int main(){
    int n;
    cin >> n;
    mtl a(n, vl(n, 0));
    ll total = 0;
    rep(i, n){
        rep(j, i) cin >> a[i][j], total += a[i][j];
        repl(j, i + 1, n) cin >> a[i][j], total += a[i][j];
        reverse(all(a[i]));
    }
    reverse(all(a));
    mtl x(n + 1, vl(n + 1, 0)), y(n + 1, vl(n + 1, 0));
    repl(i, 1, n+1){
        repl(j, 1, i){
            x[i][j] = x[i][j - 1] + a[i - 1][j - 1];
        }
        per(j, i - 1){
            y[i][j] = y[i][j + 1] + a[j][i - 1];
        }
    }
    mtl dp(n + 1, vl(n + 1, 0));
    repl(i, 1, n + 1){
        dp[0][i] = dp[0][i - 1] + y[i][0];
    }
    rep(i, n+1){
        repl(j, i + 1, n+1){
            ll d = 0;
            repl(k, j + 1, n + 1){
                d += x[k][j] + y[k][i];
                dp[j][k] = max(dp[j][k], dp[i][j] + d);
            }
        }
    }
    ll ans = 0;
    rep(i, n) ans = max(ans, dp[i][n]);
    cout << total - ans << endl;
    return 0;
}
