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 rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){
	return o<<"("<<p.fs<<","<<p.sc<<")";
}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){
	o<<"{";
	for(const T& v:vc) o<<v<<",";
	o<<"}";
	return o;
}
using ll = long long;
template<class T> using V = vector<T>;
template<class T> using VV = vector<vector<T>>;
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n-1); }

#ifdef LOCAL
#define show(x) cerr << "LINE" << __LINE__ << " : " << #x << " = " << (x) << endl
#else
#define show(x) true
#endif

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);		//DON'T USE scanf/printf/puts !!
	cout << fixed << setprecision(20);

	int N;
	cin >> N;
	VV<ll> A(N,V<ll>(N));
	
	rep(i,N) rep(j,N) if(i!=j){
		cin >> A[i][j];
	}
	

	const ll inf = 1e18;
	VV<ll> dp(N,V<ll>(N,inf)),ndp(N,V<ll>(N,inf));
	dp[0][0] = 0;
	rep1(i,N-1){
		V<ll> Rs(N+1),Ls(N+1);
		rep1(x,N) Rs[x] = Rs[x-1] + A[x-1][i];
		rep1(x,N) Ls[x] = Ls[x-1] + A[i][x-1];
		auto R = [&](int l,int r,int i){	//[l,r) -> i
			return Rs[r] - Rs[l];
		};
		auto L = [&](int l,int r,int i){	//[l,r) <- i
			return Ls[r] - Ls[l];
		};
		rep(j,N) rep(k,N) ndp[j][k] = inf;
		rep(j,i) rep(k,max(j,1)){
			chmin(ndp[j][k],dp[j][k] + R(j,i,i) + L(0,k,i));
			chmin(ndp[i][j],dp[j][k] + L(0,j,i));
		}
		swap(dp,ndp);
	}
	ll ans = inf;
	rep(j,N) rep(k,N) chmin(ans,dp[j][k]);
	cout << ans << endl;
}
