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>
#define LL long long
using namespace std;
inline int read(){
    int x=0; bool f=1; char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=0;
    for(; isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^'0');
    if(f) return x;
    return -x;
}
const int N = 500;
LL a[N+3][N+3],s[2][N+3][N+3],f[N+3][N+3];
int n;
void update(LL &x,LL y) {x = x<y?x:y;}
LL sum(int t,int lx,int rx,int ly,int ry){
    return s[t][rx][ry]-s[t][lx-1][ry]-s[t][rx][ly-1]+s[t][lx-1][ly-1];
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        for(int j=1; j<=n; j++){
            if(j==i) continue;
            scanf("%lld",&a[i][j]);
        }
    }
    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(f,42,sizeof(f)); 
    f[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=f[i][j]+sum(1,k+1,n,j+1,i)+sum(0,i+1,k,i+1,k);
                update(f[k][i],tmp);
            }
    LL ans=f[n][1];
    for(int i=1;i<=n;i++)update(ans,f[n][i]);
    printf("%lld\n",ans);
    return 0;
}
