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
#pragma GCC optimize("2,Ofast,inline")
#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
const LL N = 6e4 + 10;
const LL inf = 0x3f3f3f3f;
const LL mod = 998244353;
using namespace std;
template <typename tp>
inline void read(tp &x)
{
    x = 0; char c = getchar(); bool f = 0;
    for (; c < '0' || c > '9'; f |= (c == '-'), c = getchar()) ;
    for (; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + c - '0', c = getchar()) ;
    if (f) x = -x;
}
LL n,cost[510][510];
LL along[510][510],sum_along[510][510];
LL inverse[510][510],sum_inverse[510][510];
LL dp[510][510];
signed main()
{
    read(n);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            if(i==j) continue;
            read(cost[i][j]);
        }
    for(int i=1;i<=n;i++)
        for(int j=1;j<i;j++)
            sum_along[j][i]=sum_along[j-1][i]+cost[j][i];
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n+1;j++)
            along[i][j]=along[i][j-1]+sum_along[j-1][j]-sum_along[i-1][j];
    for(int i=1;i<=n;i++)
        for(int j=1;j<i;j++)
            sum_inverse[j][i]=sum_inverse[j-1][i]+cost[i][j];
    for(int i=1;i<=n+1;i++)
        for(int j=n+1;j>i;j--)
            inverse[i][j]=inverse[i][j+1]+sum_inverse[i][j];
    memset(dp,0x3f,sizeof(dp));
    dp[0][0]=0;
    for(int i=0;i<=n;i++)
        for(int j=i;j<=n;j++)
            if(dp[i][j]<1e18)
                for(int k=j+1;k<=n+1;k++)
                    dp[j][k]=min(dp[j][k],dp[i][j]+along[j+1][k]+inverse[j][k+1]-inverse[i][k+1]);
    LL ans=1e18;
    for(int i=0;i<=n;i++)
        ans=min(ans,dp[i][n+1]);
    printf("%lld\n",ans);
    return 0;
}
