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 pb          emplace_back
#define ll          long long
#define fi          first
#define se          second
#define mp          make_pair
#define sz(x)       int(x.size())
#define int         int64_t

using namespace std;

typedef pair<int, int> pii;
const int N = 5e2 + 5;
const int inf = (ll)1e18;

int f[N][N], n, c[N][N], cost[N][N];

int get(int x, int y, int u, int v) {
    if(x > u || y > v) return 0;
    return c[u][v] + c[x - 1][y - 1] - c[u][y - 1] - c[x - 1][v];
}

int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    #define Task        "test"
    if(fopen(Task".inp", "r")) {
        freopen(Task".inp", "r", stdin);
        freopen(Task".out", "w", stdout);
    }
    cin >> n;
    for(int i = 1; i <= n + 1; ++i) {
        for(int j = 1; j <= n + 1; ++j) {
            if(i != j && i <= n && j <= n) cin >> c[i][j];
            c[i][j] += c[i - 1][j] + c[i][j - 1] - c[i - 1][j - 1];
        }
    }
    /// cost(i, j) = sum c[a][b] | i < a < b <= j
    for(int i = 0; i <= n + 1; ++i) {
        for(int j = i + 1; j <= n + 1; ++j)
            cost[i][j] = cost[i][j - 1] + get(i + 1, j, j - 1, j);
    }
    fill_n(&f[0][0], N * N, inf);
    for(int i = 1; i <= n + 1; ++i) f[0][i] = cost[0][i];
    auto minimize = [&](int& x, int y) {if(x > y) x = y;};
    for(int i = 0; i <= n; ++i) {
        for(int j = i + 1; j <= n + 1; ++j) {
            if(f[i][j] == inf) continue;
            for(int k = j + 1; k <= n + 1; ++k) {
                minimize(f[j][k], f[i][j] + cost[j][k] + get(k + 1, i + 1, n + 1, j));
            }
        }
    }
    int res = inf;
    for(int i = 0; i <= n; ++i) res = min(res, f[i][n + 1]);
    cout << res;
}
