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 <cstdio>
#include <cstring>

template<typename T>
inline int chkmin(T &x, T y) {
  return x > y ? x = y, 1 : 0;
}

const int MAXN = 500;

int n;
int A[MAXN + 9][MAXN + 9];

long long sum[2][MAXN + 9][MAXN + 9];

void Calc_Sum() {
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      sum[0][i][j] += sum[0][i][j - 1];
      sum[1][i][j] += sum[1][i][j - 1];
    }
    for (int j = 1; j <= n; ++j) {
      sum[0][i][j] += sum[0][i - 1][j];
      sum[1][i][j] += sum[1][i - 1][j];
    }
  }
}

long long dp[MAXN + 9][MAXN + 9];

inline long long cost(int opt, int lx, int rx, int ly, int ry) {
  return sum[opt][rx][ry] - sum[opt][lx - 1][ry] - sum[opt][rx][ly - 1] + sum[opt][lx - 1][ly - 1];
}

int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) 
    for (int j = 1; j <= n; ++j)
      if (i != j) {
        scanf("%d", &A[i][j]);
        i < j ? sum[0][i][j] = A[i][j] : sum[1][i][j] = A[i][j]; 
      }

  Calc_Sum();

  memset(dp, 0x3f, sizeof dp);
  dp[0][0] = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < (!i ? 1 : i); ++j)
      for (int k = i + 1; k <= n; ++k)
        chkmin(dp[k][i], dp[i][j] + cost(1, k + 1, n, j + 1, i) + cost(0, i + 1, k, i + 1, k));

  long long ans = 0x3f3f3f3f3f3f3f3fLL;
  for (int i = 0; i <= n; ++i) chkmin(ans, dp[n][i]);

  printf("%lld\n", ans);

  return 0;
}