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 rep(i,a,b) for (int i = (a); i <= (b); ++ i)
#define rrp(i,a,b) for (int i = (a); i >= (b); -- i)
#define gc() getchar()
#define fir first
#define sec second
typedef pair<int,int> pii;
typedef double db;
typedef long long ll;
template <typename tp>
inline void read(tp& x) {
  x = 0; char tmp; bool key = 0;
  for (tmp = gc(); !isdigit(tmp); tmp = gc())
    key = (tmp == '-');
  for (; isdigit(tmp); tmp = gc()) 
    x = (x << 3) + (x << 1) + (tmp ^ '0');
  if (key) x = -x;
}
template <typename tp>
inline void ckmn(tp& x,tp y) {
  x = x < y ? x : y;
}
template <typename tp>
inline void ckmx(tp& x,tp y) {
  x = x < y ? y : x;
}

const int N = 510;
ll dp[N][N], n;
ll mat[N][N], sum[2][N][N];
ll calc(int tp,int l1,int r1,int l2,int r2) {
  return sum[tp][r1][r2] - sum[tp][l1-1][r2] - sum[tp][r1][l2-1] + sum[tp][l1-1][l2-1];
}
int main() {
  read(n);
  rep (i, 1, n) {
    rep (j, 1, n) if (i != j) {
      read(mat[i][j]);
    }
  }
  rep (i, 1, n) rep (j, 1, n)
    sum[i < j][i][j] = mat[i][j];
  rep (k, 0, 1) {
    rep (i, 1, n) rep (j, 1, n)
      sum[k][i][j] += sum[k][i][j-1];
    rep (i, 1, n) rep (j, 1, n)
      sum[k][i][j] += sum[k][i-1][j];
  }
  memset(dp, 0, sizeof dp);
  rep (i, 1, n) rep (j, 1, n) dp[0][0] += mat[i][j];
  // 1 >= qj + .. + q_{i-1} (i > j)
  rep (i, 1, n) {
    rep (a, 0, i-1) rep (b, 0, max(0, a-1)) {
      ckmx(dp[i][a], dp[a][b] - calc(1, a+1, i, a+1, i) - calc(0, i+1, n, b+1, a));
    }
  }
  ll ans = 0;
  rep (a, 0, n) rep (b, 0, max(0, a-1))
    ckmx(ans, dp[a][b] - calc(1, a+1, n, a+1, n));
  ans = dp[0][0] - ans;
  cout << ans << endl;
  return 0;
}
