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>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

const ll INF = 1e18;
const int MAXN = 505;

int n;
ll w[MAXN][MAXN],f[MAXN][MAXN],sum1[MAXN][MAXN],sum2[MAXN][MAXN];

ll getVal1(int a,int b,int c,int d){
	return sum1[b][d] - (c == 0 ? 0 : sum1[b][c - 1]) - (a == 0 ? 0 : sum1[a - 1][d]) + (a == 0 || c == 0 ? 0 : sum1[a - 1][c - 1]);
}

ll getVal2(int a,int b,int c,int d){
	return sum2[b][d] - (c == 0 ? 0 : sum2[b][c - 1]) - (a == 0 ? 0 : sum2[a - 1][d]) + (a == 0 || c == 0 ? 0 : sum2[a - 1][c - 1]);
}

int main(){
    //freopen(".in","r",stdin);
    //freopen(".out","w",stdout);
	scanf("%d",&n);
	for(int i = 1;i <= n;i++){
		for(int j = 1;j <= n;j++){
			if(i != j)
			    scanf("%d",&w[i][j]);
		}
	}
	for(int i = 1;i <= n;i++){
		for(int j = 1;j <= n;j++)
		    sum1[i][j] = sum1[i - 1][j] + sum1[i][j - 1] - sum1[i - 1][j - 1] + w[i][j];
	}
	for(int i = 1;i <= n;i++){
		for(int j = 1;j <= n;j++)
		    sum2[i][j] = sum2[i - 1][j] + sum2[i][j - 1] - sum2[i - 1][j - 1] + (i < j ? w[i][j] : 0);
	}
	for(int i = 1;i <= n;i++){
		f[i][1] = getVal2(0,i,0,i);
		for(int j = 2;j <= i;j++){
			f[i][j] = INF;
			for(int k = 1;k < j;k++)
				f[i][j] = min(f[i][j],f[j - 1][k] + getVal2(j,i,j,i) + getVal1(j,i,0,k - 1));
		}
	}
	ll ans = INF;
	for(int i = 1;i <= n;i++)
	    ans = min(ans,f[n][i]);
	printf("%lld\n",ans);
	return 0;
}
