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;

typedef long long ll;
typedef unsigned int ui;
typedef pair<int, int> pi;

#define fi first
#define se second

int read() {
  int res = 0, w = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') c = getchar(), w = -1;
  while (isdigit(c)) res = (res << 1) + (res << 3) + c - '0', c = getchar();
  return res * w;
}

const int N = 510;
ll a[N][N];
ll s[2][N][N];
ll dp[N][N];
int n;

void ck(ll &x, ll y) { x = x < y ? x : y; }

ll getsum(int typ, int lx, int rx, int ly, int ry) {
  return s[typ][rx][ry] - s[typ][lx - 1][ry] - s[typ][rx][ly - 1] + s[typ][lx - 1][ly - 1];
}

int main() {
  n = read();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) if (i != j) a[i][j] = read();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i < j) s[0][i][j] = a[i][j];
      s[0][i][j] += s[0][i][j - 1];
    }
    for (int j = 1; j <= n; j++) s[0][i][j] += s[0][i - 1][j];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i > j) s[1][i][j] = a[i][j];
      s[1][i][j] += s[1][i][j - 1];
    }
    for (int j = 1; j <= n; j++) s[1][i][j] += s[1][i - 1][j];
  }
  memset(dp, 0x3f, sizeof(dp));
  dp[0][0] = 0ll;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < max(i, 1); j++) {
      for (int k = i + 1; k <= n; k++) {
        ll tmp = dp[i][j] + getsum(1, k + 1, n, j + 1, i) + getsum(0, i + 1, k, i + 1, k);
        ck(dp[k][i], tmp);
      }
    }
  }
  ll ans = dp[n][1];
  for (int i = 1; i <= n; i++) ck(ans, dp[n][i]);
  printf("%lld\n", ans);
  return 0;
}