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;
int read();

int n;
long long f[1003][1003], s[2][1003][1003];
long long qry0(int i, int j) { return s[0][n][i] - s[0][j][i]; }
long long qry1(int i, int j, int k) {
    return s[1][n][j] - s[1][n][k] - s[1][i][j] + s[1][i][k];
}
int main() {
    n = read();
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j)
            if (i < j)
                s[0][i][j] = read();
            else if (i > j)
                s[1][i][j] = read();

    n += 2;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j) {
            s[0][i][j] += s[0][i][j - 1] + s[0][i - 1][j] - s[0][i - 1][j - 1];
            s[1][i][j] += s[1][i][j - 1] + s[1][i - 1][j] - s[1][i - 1][j - 1];
        }

    for (int i = 0; i <= n; ++i) f[i][0] = qry0(i, 0);
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j < i; ++j) {
            f[i][j] = 10000000000000000ll;
            for (int k = 0; k < j; ++k)
                f[i][j] = min(f[i][j], f[j][k] + qry1(i, j, k));
            f[i][j] += qry0(i, j);
        }

    printf("%lld\n", f[n][n - 1]);
    return 0;
}
const int _SIZE = 1 << 22;
char ibuf[_SIZE], *iS = ibuf, *iT = ibuf;
#define gc                                                         \
    (iS == iT ? iT = ((iS = ibuf) + fread(ibuf, 1, _SIZE, stdin)), \
     (iS == iT ? EOF : *iS++) : *iS++)
int read() {
    int x = 0, f = 1;
    char c = gc;
    while (!isdigit(c)) f = (c == '-') ? -1 : f, c = gc;
    while (isdigit(c)) x = x * 10 + c - '0', c = gc;
    return x * f;
}