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 <algorithm>  
#include <iostream>  
#include <sstream>  
#include <string>  
#include <cstring>
#include <vector>  
#include <queue>  
#include <set>  
#include <map>  
#include <cstdio>  
#include <cstdlib>  
#include <cctype>  
#include <cmath>  
#include <list>  
#include <cassert>
#include <ctime>
#include <climits>
using namespace std;  

#define PB push_back  
#define MP make_pair  
#define SZ(v) ((int)(v).size())  
#define FOR(i,a,b) for(int i=(a);i<(b);++i)  
#define REP(i,n) FOR(i,0,n)  
#define FORE(i,a,b) for(int i=(a);i<=(b);++i)  
#define REPE(i,n) FORE(i,0,n)  
#define FORSZ(i,a,v) FOR(i,a,SZ(v))  
#define REPSZ(i,v) REP(i,SZ(v))  
typedef long long ll;
typedef unsigned long long ull;
ll gcd(ll a,ll b) { return b==0?a:gcd(b,a%b); }

const int MAXN=500;

int n;
int cost[MAXN][MAXN];

ll dp[MAXN][MAXN]; // dp[i][j]=minimum cost when last group is [i..j]

ll sumforwardcost[MAXN][MAXN];
ll sumbackwardcost[MAXN][MAXN];

ll solve() {
	REP(i,n) FOR(j,i,n) { sumforwardcost[i][j]=0; FOR(k,i,j) sumforwardcost[i][j]+=cost[k][j]; if(i<j) sumforwardcost[i][j]+=sumforwardcost[i][j-1]; }
	REP(i,n) FOR(j,i,n) { sumbackwardcost[i][j]=0; FOR(k,i,j) sumbackwardcost[i][j]+=cost[j][k]; if(i<j) sumbackwardcost[i][j]+=sumbackwardcost[i][j-1]; }

	REP(i,n) REP(j,n) dp[i][j]=LLONG_MAX;
	REP(j,n) dp[0][j]=sumforwardcost[0][j];
	REP(i,n) FOR(j,i,n) if(dp[i][j]!=LLONG_MAX) {
		FOR(k,j+1,n) dp[j+1][k]=min(dp[j+1][k],dp[i][j]+sumforwardcost[j+1][k]+sumbackwardcost[i][n-1]-sumbackwardcost[i][k]-sumbackwardcost[j+1][n-1]+sumbackwardcost[j+1][k]);
	}
	ll ret=LLONG_MAX; REP(i,n) ret=min(ret,dp[i][n-1]); return ret;
}

void run() {
	scanf("%d",&n);
	REP(i,n) REP(j,n) if(i==j) cost[i][j]=0; else scanf("%d",&cost[i][j]);
	printf("%lld\n",solve());
}

int main() {
	run();
	return 0;
}
