CodeForces - 965E Short Code

Discription

Arkady's code contains nn variables. Each variable has a unique name consisting of lowercase English letters only. One day Arkady decided to shorten his code.

He wants to replace each variable name with its non-empty prefix so that these new names are still unique (however, a new name of some variable can coincide with some old name of another or same variable). Among such possibilities he wants to find the way with the smallest possible total length of the new names.

A string aa is a prefix of a string bb if you can delete some (possibly none) characters from the end of bb and obtain aa.

Please find this minimum possible total length of new names.

Input

The first line contains a single integer nn (1≤n≤1051≤n≤105) — the number of variables.

The next nn lines contain variable names, one per line. Each name is non-empty and contains only lowercase English letters. The total length of these strings is not greater than 105105. The variable names are distinct.

Output

Print a single integer — the minimum possible total length of new variable names.

Examples

Input

3
codeforces
codehorses
code

Output

6

Input

5
abba
abb
ab
aa
aacada

Output

11

Input

3
telegram
digital
resistance

Output

3

Note

In the first example one of the best options is to shorten the names in the given order as "cod", "co", "c".

In the second example we can shorten the last name to "aac" and the first name to "a" without changing the other names.

 

 

   很显然是可以在trie树上贪心的。

    每个点x开一个map，存单词结尾在x子树内的点的深度集合。

    如果点x没有被单词结尾覆盖，那么就把最深的放到这里。

 

    向上合并的时候启发式合并就好啦。。。。

 

    复杂度 O(N * log ^ 2(N))

 

#include<bits/stdc++.h>
#include <typeinfo> 
#define ll long long
using namespace std;
const int maxn=100005;

map<int,int> mmp[maxn];
map<int,int> :: iterator it;
int ch[maxn][26],cnt,n,N,ans,dep[maxn],word[maxn];
char S[maxn];

inline void ins(){
	int now=0,c,depth=0;
	for(int i=0;i<N;i++){
		c=S[i]-'a';
		if(!ch[now][c]) ch[now][c]=++cnt;
		now=ch[now][c],dep[now]=++depth;
	}
	
	word[now]++;
}

void dfs(int x){
	for(int i=0,to;i<26;i++) if(ch[x][i]){
		to=ch[x][i],dfs(to);
		
		if(mmp[to].size()>mmp[x].size()) swap(mmp[to],mmp[x]);
		
		for(it=mmp[to].begin();it!=mmp[to].end();++it) mmp[x][it->first]+=it->second;
		    
		mmp[to].clear();
	}
	
	if(word[x]) mmp[x][dep[x]]+=word[x];
	else if(x){
		it=--mmp[x].lower_bound(1e9);

		ans-=(it->first)-dep[x];
		
		if(!(--(it->second))) mmp[x].erase(it);
		
		mmp[x][dep[x]]++;
	}
}


int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		fill(S,S+N,0);
		scanf("%s",S),N=strlen(S),ans+=N;
		ins();
	}
	
	dfs(0);
	
	printf("%d\n",ans);
	return 0;
}