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 N      510
// #define inf    0x7f7f7f7f
#define int    long long
using namespace std;
int a[N][N] = {};
int read()
{
    int  s = 0, w = 1;
    char c = getchar();
    while ((c < '0' || c > '9') && c != '-')
        c = getchar();
    if (c == '-')
        c = getchar(), w = -1;
    while (c <= '9' && c >= '0')
        s = (s << 3) + (s << 1) + c - '0', c = getchar();
    return s * w;
}
int n;
int b[N][N] = {}, c[N][N] = {};
int f[N][N] = {};
signed 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 = n; j >= i + 1; --j)
        {
            b[i][j] = b[i][j + 1];
            for (int k = 1; k <= i; ++k)
                b[i][j] += a[j][k];
        }
    for (int i = 1; i <= n; ++i)
        for (int j = i + 1; j <= n + 1; ++j)
        {
            c[i][j] = c[i][j - 1];
            for (int k = i; k <= j - 1; ++k)
                c[i][j] += a[k][j];
        }
    memset(f, 0x7f, sizeof(f));
    f[0][0] = 0;
    for (int i = 0; i <= n; ++i)
        for (int j = i; j <= n; ++j)
            if (f[i][j] < 1e18)
                for (int k = j + 1; k <= n + 1; ++k)
                    f[j][k] = min(f[j][k], f[i][j] + c[j + 1][k] + b[j][k + 1] - b[i][k + 1]);
    int minn = 1e18;
    for (int i = 0; i <= n; ++i)
        minn = min(minn, f[i][n + 1]);
    printf("%lld\n", minn);
    return 0;
}
