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 N 510
#define ll long long
#define For(i,x,y) for(int i=(x);i<=(y);++i)
#define Rof(i,x,y) for(int i=(x);i>=(y);--i)
#define Edge(x) for(int i=head[x];i;i=e[i].nxt)
#define mset(x,y) memset(x,y,sizeof(x))
using namespace std;
ll dp[N][N],A[N][N],B[N][N],a[N][N],b[N][N];
ll getB(int rx,int ry,int lx,int ly){
	if(lx>rx||ly>ry) return 0;
	return B[rx][ry]-B[lx-1][ry]-B[rx][ly-1]+B[lx-1][ly-1];
}
ll getA(int rx,int ry,int lx,int ly){
	if(lx>rx||ly>ry) return 0;
	return A[rx][ry]-A[lx-1][ry]-A[rx][ly-1]+A[lx-1][ly-1];
}
void chkmin(ll &x,ll y){ if(y<x) x=y; }
int main(){
	int n;ll ans=0;
	scanf("%d",&n);
	For(i,1,n){
		For(j,1,i-1) scanf("%lld",&b[i][j]);
		For(j,i+1,n) scanf("%lld",&a[i][j]);
		For(j,1,n){
			A[i][j]=A[i-1][j]+A[i][j-1]+a[i][j]-A[i-1][j-1];
			B[i][j]=B[i-1][j]+B[i][j-1]+b[i][j]-B[i-1][j-1];
		}
	}
	mset(dp,0x3f);ans=dp[0][0];
	dp[0][0]=0;
	For(i,0,n) For(j,0,max(i-1,0)) For(k,i+1,n)
		chkmin(dp[k][i],dp[i][j]+getB(n,i,k+1,j+1)+getA(k,k,i+1,i+1));
	For(j,1,n) chkmin(ans,dp[n][j]);
	printf("%lld",ans);
}