It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of the cities connected. Given the map of cities which have all the remaining highways marked, you are supposed to tell the number of highways need to be repaired, quickly.

For example, if we have 3 cities and 2 highways connecting city​1​​-city​2​​ and city​1​​-city​3​​. Then if city​1​​ is occupied by the enemy, we must have 1 highway repaired, that is the highway city​2​​-city​3​​.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 3 numbers N (<1000), M and K, which are the total number of cities, the number of remaining highways, and the number of cities to be checked, respectively. Then M lines follow, each describes a highway by 2 integers, which are the numbers of the cities the highway connects. The cities are numbered from 1 to N. Finally there is a line containing K numbers, which represent the cities we concern.

Output Specification:

For each of the K cities, output in a line the number of highways need to be repaired if that city is lost.

Sample Input:

3 2 3
1 2
1 3
1 2 3

Sample Output:

1
0
0

1. 題目裡雖然說了“It is vitally important to have all the cities connected by highways in a war.”，但他給你的圖不一定是一張連通圖，甚至還會有孤立頂點，你得把這些圖全都連起來
2. 用動態申請記憶體的鄰接表會超時，鄰接矩陣或靜態鄰接表不會超時
3. 三種思路（假設均使用靜態鄰接表）：去掉一個頂點並且對剩下的圖進行dfs（T=O(k(n+m))，很慢）；去掉一個頂點然後算並查集（T=O(k*m)）；對各子圖本別計算其關節點（T=O(n+m)）並在計算過程中順便算出其劃分出的子圖數量。不過本題k比較小，前兩個方法能過。