문제

Bessie has a simple undirected graph with vertices labeled $1\dots N$ ($2\le N\le 750$). She generates a depth-first search (DFS) order of the graph by calling the function $\texttt{dfs}(1)$, defined by the following C++ code. Each adjacency list ($\texttt{adj}[i]$ for all $1\le i\le N$) may be permuted arbitrarily before starting the depth first search, so a graph can have multiple possible DFS orders.

vector vis(N + 1); vector<vector> adj(N + 1); // adjacency list vector dfs_order;

void dfs(int x) { if (vis[x]) return; vis[x] = true; dfs_order.push_back(x); for (int y : adj[x]) dfs(y); }

You are given the initial state of the graph as well as the cost to change the state of each edge. Specifically, for every pair of vertices $(i,j)$ satisfying $1\le i<j\le N$, you are given an integer $a_{i,j}$ ($0<|a_{i,j}|\le 1000$) such that

If $a_{i,j}>0$, edge $(i,j)$ is not currently in the graph, and can be added for cost $a_{i,j}$.If $a_{i,j}<0$, edge $(i,j)$ is currently in the graph, and can be removed for cost $-a_{i,j}$.

Determine the minimum total cost to change the graph so that $[1,2\dots,N]$ is a possible DFS ordering.

입력

The first line contains $N$.

Then $N-1$ lines follow. The $j-1$th line contains $a_{1,j}, a_{2,j}, \dots, a_{j-1,j}$ separated by spaces.

출력

The minimum cost to change the graph so that $[1,2,\dots, N]$ is a possible DFS ordering.

예제 입력 1

4
1
2 3
40 6 11

예제 출력 1

10

예제 입력 2

5
-1
10 -2
10 -7 10
-6 -4 -5 10

예제 출력 2

5

예제 입력 3

4
-1
-2 300
4 -5 6

예제 출력 3

9

점수

Inputs 4-9: All $a_{i,j}>0$Inputs 10-16: $N\le 50$Inputs 17-23: No additional constraints.