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;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()
#define dump(x) cout << #x << " = " << (x) << endl
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n-1); }

template<class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
	os<<"{";
	rep(i, v.size()) {
		if (i) os<<",";
		os<<v[i];
	}
	os<<"}";
	return os;
}

inline void chmin(ll &x, ll y) { if (x > y) x = y; }

const int maxn = 512;
const ll INF = TEN(18);

ll A[maxn][maxn];
ll dp[maxn][maxn];
ll s[2][maxn][maxn];

ll get(int tp, int a, int b, int c, int d) { //sum a <= x <= c, b <= y <= d
	return s[tp][c+1][d+1] - s[tp][c+1][b] - s[tp][a][d+1] + s[tp][a][b];
}

int main() {
	int N; cin >> N;
	rep(i, N) {
		rep(j, N) {
			if (i != j) cin >> A[i][j];
			int tp = (i > j);
			rep(t, 2) {
				s[t][i+1][j+1] = s[t][i+1][j] + s[t][i][j+1] - s[t][i][j];
				if (t == tp) {
					s[t][i+1][j+1] += A[i][j];
				}
			}
		}
	}

	rep(i, maxn) rep(j, maxn) dp[i][j] = INF;
	
	for (int i = 1; i < N; ++i) {
		dp[i][0] = get(0, 0, 0, i-1, i-1);
	}

	for (int i = 1; i < N; ++i) {
		for (int j = 1; j < i; ++j) {
			rep(k, j) {
				chmin(dp[i][j], dp[j][k] + get(0, j, j, i-1, i-1) + get(1, i, k, N-1, j-1));
			}
		}
	}

	ll ans = s[0][N][N];

	for (int i = 1; i < N; ++i) {
		rep(j, i) {
			chmin(ans, dp[i][j] + get(0, i, i, N-1, N-1));
		}
	}

	cout << ans << endl;

	return 0;
}