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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>
#include <unordered_set>
#include <bitset>
#include <time.h>
#include <limits.h>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define Fi first
#define Se second
#define pb push_back
#define szz(x) (int)x.size()
#define rep(i,n) for(int i=0;i<n;i++)
#define all(x) x.begin(),x.end()
typedef tuple<int, int, int> t3;

int N, A[510][510];
ll S1[510][510], S2[510][510];
ll dp[510][510];

int main() {
	scanf("%d", &N);
	for(int i=1;i<=N;i++) {
		for(int j=1;j<=N;j++) if(i != j) scanf("%d", A[i] + j);
	}
	for(int i=1;i<=N;i++) {
		S1[i][i] = 0;
		for(int j=i+1;j<=N;j++) {
			S1[i][j] = S1[i][j-1];
			for(int k=i;k<j;k++) {
				S1[i][j] += A[k][j];
			}
		}
	}
	for(int i=1;i<=N;i++) {
		for(int j=1;j<i;j++) {
			S2[i][j] = S2[i][j-1] + A[i][j];
		}
	}
	rep(i, 510) rep(j, 510) dp[i][j] = 1e18;
	for(int i=1;i<=N;i++) {
		dp[1][i] = S1[1][i];
		for(int k=1;k<i;k++) {
			ll sum = 0;
			for(int j=i;j>k;j--) {
				sum += S2[j][k-1];
				dp[j][i] = min(dp[j][i], dp[k][j-1] + S1[j][i] + sum);
			}
		}
	}
	ll ans = 1e18;
	for(int i=1;i<=N;i++) ans = min(ans, dp[i][N]);
	
	printf("%lld\n", ans);
	return 0;
}