CodeForces

Greg and Graph

Greg has a weighed directed graph, consisting of n vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:

• The game consists of n steps.
• On the i-th step Greg removes vertex numberx_i from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex.
• Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume thatd(i, v, u) is the shortest path between verticesv and u in the graph that formed before deleting vertexx_i, then Greg wants to know the value of the following sum:.

Help Greg, print the value of the required sum before

Input

The first line contains integer n (1 ≤ n ≤ 500) — the number of vertices in the graph.

Next n lines contain n integers each — the graph adjacency matrix: the j-th number in the i-th linea_ij(1 ≤ a_ij ≤ 10⁵, a_ii = 0) represents the weight of the edge that goes from vertex i to vertexj.

The next line contains n distinct integers:x₁, x₂, ..., x_n(1 ≤ x_i ≤ n) — the vertices that Greg deletes.

Output

Print n integers — the i-th number equals the required sum before the i-th step.

Please, do not use the %lld specifier to read or write 64-bit integers inC++. It is preferred to use the cin, cout streams of the %I64d specifier.

Example

Input

101

Output

0 

Input    
20 54 01 2
Output    
9 0 
Input    
40 3 1 16 0 400 12 4 0 11 1 1 04 1 2 3
Output    
17 23 404 0 
题意：有一张有向图，Greg玩一个游戏，游戏有n步，每步是删除一个点，包括和这个点相连的所有边，求出每删除一个点之前的任意两个点的最短路的和
思路：求这种任意两点之间的最短路，一般都会想到Floyd-Warshall算法，关键是怎么用，这个算法的核心就是那个三重循环，最外层那个循环每次增加一个点，然后进行松弛操作，第一次是是否能通过1这个点进行松弛，第二次是是否通过1，2这两个点进行松弛，有点动态规划的意思，那么对于这个题我们可以用一个逆向思维，我们从最后一个一个的加点，这就和Fld算法一样了，只是在求最短距离和的时候我们只加那些已被我们加入的点的最短和
#include<cstdio>#include<algorithm>#include<iostream>using namespace std;typedef long long ll;const ll INF=0x3f3f3f3f;const int maxn=505;ll e[maxn][maxn],ans[maxn];int a[maxn];bool vis[maxn];int n;ll sum;void init(){for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++){if(i==j)               e[i][j]=0;            else e[i][j]=INF;  }}int main(void){scanf("%d",&n);init();for(int i=1;i<=n;i++)   for(int j=1;j<=n;j++)      scanf("%d",&e[i][j]);for(int i=1;i<=n;i++)      scanf("%d",&a[i]);for(int k=n;k>=1;k--){sum=0;        vis[a[k]]=true;        for(int i=1;i<=n;i++)           for(int j=1;j<=n;j++)      e[i][j]=min(e[i][j],e[i][a[k]]+e[a[k]][j]);for(int i=1;i<=n;i++)   for(int j=1;j<=n;j++)      if(e[i][j]!=INF&&vis[i]&&vis[j])     sum+=e[i][j];ans[k]=sum;    }  for(int i=1;i<=n;i++)   cout<<ans[i]<<" ";cout<<endl;return 0;      }