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 all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
const int maxn = 505;
ll n, dp[maxn][maxn], a[maxn][maxn], b[maxn][maxn];
ll get(ll i, ll l, ll r) {
	return r >= l ? a[i][r] - a[i][l - 1] : 0;
}
ll rget(ll i, ll l, ll r) {
	return r >= l ? b[i][r] - b[i][l - 1] : 0;
}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n;
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++) {
			if(i != j) cin >> a[i][j];
			b[j][i] = a[i][j];
		}
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++) {
			a[i][j] += a[i][j-1];
			b[i][j] += b[i][j-1];
		}
		
	memset(dp, 0x3f, sizeof dp);
	dp[0][0] = 0;
	for(int i = 0; i <= n; i++)
		for(int j = i; j <= n; j++)
			if(dp[i][j] < 1ll<<60) {
				ll cur = 0;
				for(int k = j+1; k <= n; k++) {
					cur += get(k, 1, i-1);
					cur += rget(k, j+1, k);
					dp[j+1][k] = min(dp[j+1][k], dp[i][j] + cur);
				}
			}
	ll ans = 1ll<<60;
	for(int i = 1; i <= n; i++) ans = min(ans, dp[i][n]);
	cout << ans << '\n';
	return 0;
}
