IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2) 653B Bear and Compressing(dfs + stl)

B. Bear and Compressing

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Limak is a little polar bear. Polar bears hate long strings and thus they like to compress them. You should also know that Limak is so young that he knows only first six letters of the English alphabet: 'a', 'b', 'c', 'd', 'e' and 'f'.

You are given a set of q possible operations. Limak can perform them in any order, any operation may be applied any number of times. The i-th operation is described by a string ai of length two and a string bi of length one. No two of q possible operations have the same string ai.

When Limak has a string s he can perform the i-th operation on s if the first two letters of s match a two-letter string ai. Performing the i-th operation removes first two letters of s and inserts there a string bi. See the notes section for further clarification.

You may note that performing an operation decreases the length of a string s exactly by 1. Also, for some sets of operations there may be a string that cannot be compressed any further, because the first two letters don't match any ai.

Limak wants to start with a string of length n and perform n - 1 operations to finally get a one-letter string "a". In how many ways can he choose the starting string to be able to get "a"? Remember that Limak can use only letters he knows.

Input

The first line contains two integers n and q (2 ≤ n ≤ 6, 1 ≤ q ≤ 36) — the length of the initial string and the number of available operations.

The next q lines describe the possible operations. The i-th of them contains two strings ai and bi (|ai| = 2, |bi| = 1). It's guaranteed thatai ≠ aj for i ≠ j and that all ai and bi consist of only first six lowercase English letters.

Output

Print the number of strings of length n that Limak will be able to transform to string "a" by applying only operations given in the input.

Examples

input

3 5
ab a
cc c
ca a
ee c
ff d

output

4

input

2 8
af e
dc d
cc f
bc b
da b
eb a
bb b
ff c

output

1

input

6 2
bb a
ba a

output

0

Note

In the first sample, we count initial strings of length 3 from which Limak can get a required string "a". There are 4 such strings: "abb", "cab", "cca", "eea". The first one Limak can compress using operation 1 two times (changing "ab" to a single "a"). The first operation would change "abb" to "ab" and the second operation would change "ab" to "a".

Other three strings may be compressed as follows:

• "cab"  "ab"  "a"
• "cca"  "ca"  "a"
• "eea"  "ca"  "a"

In the second sample, the only correct initial string is "eb" because it can be immediately compressed to "a".

题目链接:点击打开链接

给出n, q, 有q种字符串对应的关系, 表示两个字符可以转化为一个字符, 问有多少个长度为n 由abcdef构成的字符串, 两两转化最后得到

字符'a'.

一共才6个字符, 枚举枚举所有可能出现的情况进行判断即可.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
#include "cstdlib"
using namespace std;
typedef long long ll;
#define X first
#define Y second
#define exp 1e-8
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 10;
int n, q, sum;
std::map<string, string> mp;
string s[MAXN];
void Dfs(string x, int len)
{
	if(len >= 2 && mp[x.substr(0, 2)] == "") return;
	if(len == n) {
		string ans = x.substr(0, 2);
		if(mp[ans] != "") ans = mp[ans];
		else return;
		for(int i = 2; i < n; ++i) {
			string tmp = ans + x[i];
			if(mp[tmp] != "") ans = mp[tmp];
			else return;
		}
		if(ans == s[0]) sum++;
		return;
	}
	for(int i = 0; i < 6; ++i) 
		Dfs(x + s[i], len + 1);
	return;
}
int main(int argc, char const *argv[])
{
	for(int i = 0; i < 6; ++i)
		s[i] = 'a' + i;
	cin >> n >> q;
	while(q--) {
		string x, y;
		cin >> x >> y;
		mp[x] = y;
	}
	for(int i = 0; i < 6; ++i)
		Dfs(s[i], 1);
	cout << sum << endl;
	return 0;
}