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 long long ll;
typedef pair<int,int> pp;
typedef pair<ll,ll> pll;
void read(int& x){ scanf("%d",&x); }
void read(ll& x){ scanf("%lld",&x); }
void read(pp& x){ scanf("%d%d",&x.first, &x.second); }
void read(pll& x){ scanf("%lld%lld",&x.first, &x.second); }
template<typename T,typename... Args>
void read(T& a,Args&... b){ read(a); read(b...); }
void cppio(){ ios_base::sync_with_stdio(0); cin.tie(0); }
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define eb emplace_back
#define x first
#define y second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define sz(x) (int)(x).size()

const int maxn = 510;
int n;
int cost[maxn][maxn];

ll dp[maxn][maxn];

ll cp[maxn][maxn];
ll gp[maxn][maxn];

int main()
{
	cppio();
	cin >> n;
	rrep(i, n) rrep(j, n) if(i != j) cin >> cost[i][j];

	rrep(i, n) rrep(j, n)
		cp[i][j] = cp[i-1][j] + cp[i][j-1] - cp[i-1][j-1] + cost[i][j];

	rrep(i, n) {
		for(int j=i+1; j<=n; ++j) {
			gp[i][j] = gp[i][j-1];
			for(int k=i; k<j; ++k) {
				gp[i][j] += cost[k][j];
			}
		}
	}

	memset(dp, 0x7f, sizeof(dp));

	dp[1][1] = 0;
	for(int i=2; i<=n; ++i)
		dp[1][i] = gp[1][i];

	auto rect = [&](int l, int r, int d, int u) -> ll {
		if(l>r || d>u) return 0;
		return cp[r][u] - cp[l-1][u] - cp[r][d-1] + cp[l-1][d-1];
	};

	rrep(l, n) {
		for(int r=l; r<=n; ++r) {
			for(int R=r+1; R<=n; ++R) {
				dp[r+1][R] = min(dp[r+1][R],
					dp[l][r] + gp[r+1][R] +
						rect(r+1, R, 1, l-1));
			}
		}
	}

/*
	rrep(i, n) {
		for(int j=i; j<=n; ++j) printf("dp[%d][%d] = %lld\n", i, j, dp[i][j]);
	}
*/

	ll ans = 1ll << 60;
	rrep(i, n) ans = min(ans, dp[i][n]);

	cout << ans << endl;

	return 0;
}
