Mroning_X的字典

大家都知道Mroning_X最近打编一本acm常用单词查询词典，但是在编这本书的时候，输入的单词是没有按照字典序的，Mroning_X看着一堆混乱的单词突然他就想玩游戏了，于是他就想出了一个游戏规则：不能交换两两单词的数量，但是可以对第i个单词进行翻转操作，这操作要花费value[i]，求出最少花费多少才可以让全部单词满足字典序（str[i]<=str[i+1]），如果不行则输出-1。Mroning_X心满意足的将这题用来提问路过的ACMer，不能在1s内答出来的人会就会被Mroning_X抓去吊在505门口打。

现在轮到你了！！！！！hiahiahiahia！！！！！

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of strings.

The second line contains n integers c_i (0 ≤ c_i ≤ 10⁹), the i-th of them is equal to the amount of energy Vasiliy has to spent in order to reverse the i-th string.

Then follow n lines, each containing a string consisting of lowercase English letters. The total length of these strings doesn't exceed 100 000.

第一行是一个整数N，表示有多少个单词（2<=N<=100000）；

第二行有N个数，表示翻转第i个的value；（0<=value[i]<=1e9）

接下来n行，每行一个单词（单词的长度并没有告诉你, 但全部的单词长度加起来一定不会超过100000）

If it is impossible to reverse some of the strings such that they will be located in lexicographical order, print  - 1. Otherwise, print the minimum total amount of energy Vasiliy has to spent.

输出将所有单词字典序从小到大排列好的方案的最小花费，如果不存在将所有单词字典序从小到大排列好的方案则输出-1；

2
1 2
ba
ac

1

3
1 3 1
aa
ba
ac

1

2
5 5
bbb
aaa

-1

2
3 3
aaa
aa

-1

 Note

In the second sample one has to reverse string 2 or string 3. To amount of energy required to reverse the string 3 is smaller.第一个样例，只交换哪一个都可以，但是交换1的费用更小；

In the third sample, both strings do not change after reverse and they go in the wrong order, so the answer is  - 1。第三个样例，怎么交换都没用，而且字典序是错的.
In the fourth sample, both strings consists of characters 'a' only, but in the sorted order string "aa" should go before string "aaa", thus the answer is  - 1.第四个样例，怎么交换都没用，且aaa>aa