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,s,t) for(int i=s;i<t;i++)
typedef long long ll;
const ll inf=1e18;
ll dp[555][555],sum[2][555][555];
ll cal(int t,int l1,int r1,int l2,int r2)
{
    if(!r1||!r2)return 0;
    return sum[t][r1][r2]-sum[t][l1-1][r2]-sum[t][r1][l2-1]+sum[t][l1-1][l2-1];
}

int main()
{
    int n;scanf("%d",&n);
    rep(i,1,n+1)rep(j,1,n+1)
    if(i!=j)
    {
        scanf("%lld",&sum[0][i][j]);
        if(i<j)sum[1][i][j]=sum[0][i][j];
    }
    rep(k,0,2)rep(i,1,n+1)rep(j,1,n+1)
        sum[k][i][j]+=sum[k][i-1][j]+sum[k][i][j-1]-sum[k][i-1][j-1];
    rep(i,0,n+1)rep(j,0,n+1)dp[i][j]=inf;
    dp[0][0]=0;
    rep(i,0,n+1)
    rep(j,0,n+1)if(dp[i][j]!=inf)
    {
        rep(k,j+1,n+1)
        dp[j][k]=min(dp[j][k],dp[i][j]+cal(0,j+1,k,1,i)+cal(1,j+1,k,j+1,k));
    }
    ll res=inf;
    for(int i=n-1;i>=0;i--)res=min(res,dp[i][n]);
    printf("%lld\n",res);
}
