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 ll long long 
#define nn 508
#define mod 998244353
ll n;ll cost[nn][nn];
ll dp[nn][nn];
void upd(ll&x,ll y){
	x=min(x,y);
}
ll pre[nn][nn],pres[nn][nn];ll prec[nn][nn],precs[nn][nn];

ll cut(ll l,ll r,ll x,ll y){
	return pres[r][y]-pres[l-1][y]-pres[r][x-1]+pres[l-1][x-1];
}

ll ccut(ll l,ll r,ll x,ll y){
	return precs[r][y]-precs[l-1][y]-precs[r][x-1]+precs[l-1][x-1];
}
int main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;i++) {
		for(int j=1;j<=n;j++) if(i!=j) scanf("%lld",&cost[i][j]); 
		for(int j=1;j<=n;j++) pre[i][j]+=pre[i][j-1]+cost[i][j];
	}	
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) pres[i][j]+=pres[i-1][j]+pre[i][j];
	
	
	
	
	for(int i=1;i<=n;i++) {
		for(int j=1;j<=n;j++) cost[i][j]*=(i<j);
		for(int j=1;j<=n;j++) prec[i][j]+=prec[i][j-1]+cost[i][j];
	}	
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) precs[i][j]+=precs[i-1][j]+prec[i][j];
	
	ll ans=1e18;
	memset(dp,0x3f,sizeof dp);dp[0][0]=0;
	for(int x=0;x<=n;x++)
		for(int i=0;i<=n;i++){
			for(int j=0;j<=n;j++){
				upd(dp[x][i],dp[i][j]+cut(x+1,n,j+1,i)+ccut(i+1,x,i+1,x));
			}
			//dp[x][i]<<":"<<ans<<"\n";
			upd(ans,dp[x][i]+ccut(x+1,n,x+1,n));
		}
	
	printf("%lld\n",ans);
	
	return 0;
}