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;
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define dbv(v) cerr << #v << "="; for (auto _x : v) cerr << _x << ", "; cerr << endl
#define dba(a, n) cerr << #a << "="; for (int _i = 0; _i < (n); ++_i) cerr << a[_i] << ", "; cerr << endl
typedef long long ll;
typedef long double ld;
ll A[505][505], pref[505], suf[505], ndp[505][505], dp[505][505];
int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < n; ++j) {
			if (i == j) continue;
			scanf("%lld", &A[i][j]);
		}
	const ll INF = 1LL << 60;
	for (int a = 0; a <= n; ++a)
		for (int b = 0; b <= n; ++b)
			dp[a][b] = INF;
	dp[0][0] = 0;
	for (int i = 1; i < n; ++i) {
		suf[i] = 0;
		for (int j = i - 1; j >= 0; --j) {
			suf[j] = suf[j + 1] + A[j][i];
		}
		pref[0] = A[i][0];
		for (int j = 1; j < i; ++j) {
			pref[j] = pref[j - 1] + A[i][j];
		}
		for (int a = 0; a <= n; ++a)
			for (int b = 0; b <= n; ++b)
				ndp[a][b] = INF;
		for (int a = 0; a < i; ++a)
			for (int b = 0; b <= a; ++b) {
				ndp[a][b] = min(ndp[a][b], dp[a][b] + (b == 0 ? 0 : pref[b - 1]) + suf[a]);
				ndp[i][a] = min(ndp[i][a], dp[a][b] + (a == 0 ? 0 : pref[a - 1]));
			}
		swap(dp, ndp);
	}
	ll ans = INF;
	for (int a = 0; a < n; ++a)
		for (int b = 0; b <= a; ++b)
			ans = min(ans, dp[a][b]);
	printf("%lld\n", ans);
}
