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>
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
using namespace std;
const ll MAX_N=500,INF=1e18;
ll N,A[MAX_N][MAX_N],psm[MAX_N+1][MAX_N+1],trism[MAX_N+1][MAX_N+1],dp[MAX_N+1][MAX_N+1];
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>N;
	REP(i,N){
		REP(j,N-1){
			ll a;
			cin>>a;
			ll inx=j;
			if(j>=i){
				inx++;
			}
			A[i][inx]=a;
		}
	}
	REP(i,N){
		REP(j,N){
			psm[i+1][j+1]=A[i][j]+psm[i+1][j]+psm[i][j+1]-psm[i][j];
		}
	}
	REP(i,N+1){
		FOR(j,i+1,N+1){
			trism[i][j]=trism[i][j-1]+psm[j-1][j]-psm[i][j]-psm[j-1][j-1]+psm[i][j-1];
		}
	}
	REP(i,N+1){
		REP(j,N+1){
			dp[i][j]=INF;
		}
	}
	dp[0][0]=0;
	REP(i,N){
		FOR(j,i,N){
			FOR(x,j,N+1){
				dp[j][x]=min(dp[j][x],dp[i][j]+trism[j][x]+psm[x][i]-psm[j][i]);
			}
		}
	}
	ll ans=INF;
	REP(i,N+1){
		ans=min(ans,dp[i][N]);
	}
	REP(i,N+1){
		REP(j,N+1){
//			printf("dp[%lld][%lld]=%lld\n",i,j,dp[i][j]);
		}
	}		
	cout<<ans<<endl;
}