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
/*
 * Author       : YangDavid
 * Created Time : 2019年07月22日 星期一 11时13分09秒
 */

#include<bits/stdc++.h>
#define rep(i, n) for(int i = 1; i <= n; ++i)
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

const int maxn = 555;
const ll INF = 0x3f3f3f3f3f3f3f3f;
int n, a[maxn][maxn];
ll dp[maxn][maxn], pre1[maxn][maxn], pre2[maxn][maxn];
void upd(ll &x, ll y) { if(x > y) x = y; }

int main() {
    scanf("%d", &n);
    rep(i, n) rep(j, n) if(i != j) scanf("%d", &a[i][j]);
    rep(i, n) rep(j, n) {
        pre1[i][j] = pre1[i - 1][j] + pre1[i][j - 1] - pre1[i - 1][j - 1] + a[i][j];
        pre2[i][j] = pre2[i - 1][j] + pre2[i][j - 1] - pre2[i - 1][j - 1] + (i<j ? a[i][j] : 0);
    }

    memset(dp, 0x3f, sizeof dp);

    dp[0][0] = 0;
    for(int i = 0; i < n; ++i) {
        for(int last = 0; last <= i; ++last) {
            for(int j = i + 1; j <= n; ++j) {
                upd(dp[j][i], dp[i][last] + pre1[j][last] - pre1[i][last] + pre2[j][j] - pre2[i][j] - pre2[j][i] + pre2[i][i]); 
            }
        }
    }
    ll ans = INF;
    for(int last = 0; last < n; ++last)
        upd(ans, dp[n][last]);
    printf("%lld\n", ans);

    return 0;
}
