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#include <iostream>#include <memory.h>#include <stdlib.h>#include <string>#include <utility>#include <vector>#include <map>using namespace std;typedef vector<string> v_s;typedef pair<string,int> p_s_i;typedef map<p_s_i,int> m_s_i;v_s dict;m_s_i memo;//Trie Implementation//Complexity:	O(Length of String) look up/insertion time//Author:	Rajiv Mathews//NOTE		Use g++ -O2 for faster mapclass trie {private:	map<char, trie *> next;	bool is_word;public:	//Add given word (string) to the trie	void add_word(string s)	{		trie *curr_node = this;		for(int i = 0 ; s[i] != '\0' ; ++i) {			if(curr_node->next[s[i]] == NULL) curr_node->next[s[i]] = new trie();			curr_node = curr_node->next[s[i]];		}		curr_node->is_word = true;	}	//Check if given word (string) is present in the trie	bool is_present(string s)	{		trie *curr_node = this;		for(int i = 0 ; s[i] != 0 ; ++i) {			if(curr_node->next[s[i]] == NULL) return false;			curr_node = curr_node->next[s[i]];		}		return curr_node->is_word;	}};trie dict_words;//Count ways to split substring of 's'  in to 'm'//parts such that each part is atleast of length 'min_len'int count_ways(string s, int m, int min_len){	if(m==0) {		return s.length()==0?1:0;	}	{		p_s_i t;		if(memo.find(t=make_pair(s,m)) != memo.end()) return memo[t];	}	int ways=0;	for(int len=min_len ; len <= s.length() ; ++len) {		if(dict_words.is_present(s.substr(0,len))) {			ways += count_ways(s.substr(len,s.length()), m-1, min_len);		}	}	memo[make_pair(s,m)] = ways;	return ways;}int main(int argc, char *argv[]){	int m=atoi(argv[1]);	string temp_input;	int ans = 0;	while(cin >> temp_input) {		dict.push_back(temp_input);		dict_words.add_word(temp_input);	}	for(v_s::iterator it=dict.begin() ; it!=dict.end() ; ++it) {		ans += count_ways(*it, m, 3);	}	printf("%d\n", ans);	return 0;}