Problem:
Score : 1200 points

Problem Statement

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.

Sample Input 1

3
2 1
1 4
3 3

Sample Output 1

2

If we delete the edge (0 → 1) added by Snuke, the graph will no longer contain a negative cycle.
The cost will be 2 in this case, which is the minimum possible.

Sample Input 2

4
1 1 1
1 1 1
1 1 1
1 1 1

Sample Output 2

2

If we delete the edges (1 → 2) and (3 → 0) added by Snuke, the graph will no longer contain a negative cycle.
The cost will be 1+1=2 in this case, which is the minimum possible.

Sample Input 3

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

Sample Output 3

2280211

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define reg register
#define pr std::pair<int, int>
#define fi first
#define se second
#define FIN(s) freopen(s, "r", stdin)
#define FOUT(s) freopen(s, "w", stdout)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define rep(i, l, r) for (int i = l; i <= r; ++i)
#define lep(i, l, r) for (int i = l; i < r; ++i)
#define irep(i, r, l) for (int i = r; i >= l; --i)
#define ilep(i, r, l) for (int i = r; i > l; --i)
#define Rep(i, n) rep(i, 1, n)
#define Lep(i, n) lep(i, 1, n)
#define IRep(i, n) irep(i, n, 1)
#define ILep(i, n) ilep(i, n, 1)
typedef long long ll;
typedef long double ld;

namespace modular {
    const int MOD = 1000000007;
    inline int add(int x, int y) { return (x += y) >= MOD ? x -= MOD : x; }
    inline void inc(int &x, int y) { (x += y) >= MOD ? x -= MOD : 0; }
    inline int mul(int x, int y) { return 1LL * x * y % MOD; }
    inline int qpow(int x, int y) {
        int ans = 1;
        for (; y; y >>= 1, x = mul(x, x))
            if (y & 1) ans = mul(ans, x);
        return ans;
    }
};  // namespace modular

namespace Base {
    template <typename Tp>
    inline Tp input() {
        Tp x = 0, y = 1;
        char c = getchar();
        while ((c < '0' || '9' < c) && c != EOF) {
            if (c == '-') y = -1;
            c = getchar();
        }
        if (c == EOF) return 0;
        while ('0' <= c && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x *= y;
    }
    template <typename Tp>
    inline void read(Tp &x) {
        x = input<Tp>();
    }
    template <typename Tp>
    inline void chmax(Tp &x, Tp y) {
        x < y ? x = y : 0;
    }
    template <typename Tp>
    inline void chmin(Tp &x, Tp y) {
        x > y ? x = y : 0;
    }
};  // namespace Base
using namespace Base;
/*----------------------------------------------------------------------------*/

#define MAX_N 507

int N;
ll f[MAX_N][MAX_N], s1[MAX_N][MAX_N], s2[MAX_N][MAX_N];

inline ll calc(int k, int j, int i) {
    return s2[j + 1][i] + s1[N][j] - s1[N][k] - s1[i][j] + s1[i][k];
}

int main() {
    read(N);
    Rep(i, N) Rep(j, N) if (i != j) read(s1[i][j]);
    Rep(j, N) IRep(i, j) s2[i][j] =
        s2[i + 1][j] + s2[i][j - 1] - s2[i + 1][j - 1] + s1[i][j];
    Rep(i, N) Rep(j, N) s1[i][j] +=
        s1[i - 1][j] + s1[i][j - 1] - s1[i - 1][j - 1];
    ll ans = 1e17;
    Lep(i, N) {
        f[i][0] = calc(0, 0, i);
        chmin(ans, f[i][0] + calc(0, i, N));
        Lep(j, i) {
            f[i][j] = 1e17;
            lep(k, 0, j) { chmin(f[i][j], f[j][k] + calc(k, j, i)); }
            chmin(ans, f[i][j] + calc(j, i, N));
        }
    }
    printf("%lld\n", ans);
    return 0;
}