题干：

Sergey attends lessons of the N-ish language. Each lesson he receives a hometask. This time the task is to translate some sentence to the N-ish language. Sentences of the N-ish language can be represented as strings consisting of lowercase Latin letters without spaces or punctuation marks.

Sergey totally forgot about the task until half an hour before the next lesson and hastily scribbled something down. But then he recollected that in the last lesson he learned the grammar of N-ish. The spelling rules state that N-ish contains some "forbidden" pairs of letters: such letters can never occur in a sentence next to each other. Also, the order of the letters doesn't matter (for example, if the pair of letters "ab" is forbidden, then any occurrences of substrings "ab" and "ba" are also forbidden). Also, each pair has different letters and each letter occurs in no more than one forbidden pair.

Now Sergey wants to correct his sentence so that it doesn't contain any "forbidden" pairs of letters that stand next to each other. However, he is running out of time, so he decided to simply cross out some letters from the sentence. What smallest number of letters will he have to cross out? When a letter is crossed out, it is "removed" so that the letters to its left and right (if they existed), become neighboring. For example, if we cross out the first letter from the string "aba", we get the string "ba", and if we cross out the second letter, we get "aa".

Input

The first line contains a non-empty string s, consisting of lowercase Latin letters — that's the initial sentence in N-ish, written by Sergey. The length of string sdoesn't exceed 105.

The next line contains integer k (0 ≤ k ≤ 13) — the number of forbidden pairs of letters.

Next k lines contain descriptions of forbidden pairs of letters. Each line contains exactly two different lowercase Latin letters without separators that represent the forbidden pairs. It is guaranteed that each letter is included in no more than one pair.

Output

Print the single number — the smallest number of letters that need to be removed to get a string without any forbidden pairs of neighboring letters. Please note that the answer always exists as it is always possible to remove all letters.

Examples

Input

ababa
1
ab

Output

2

Input

codeforces
2
do
cs

Output

1

Note

In the first sample you should remove two letters b.

In the second sample you should remove the second or the third letter. The second restriction doesn't influence the solution.

题目大意：

   在一串文字中有k个字符对（2位）不允许出现，问最小修改数。输入时的字符对保证两个字符不同，且其中字符唯一

解题报告：

    一个字串只含有某个对中的字母，这个字符对才可能存在，要破坏这个字串，就在改变出现次数最少的那个字符，由此寻找符合题意的字串，并且求出修改次数，叠加即可。（主要是这题说明了，这k对字符串都不是重复字符）

既然k个两字符串中没有重复的串，那么你删除一些必要的字符后肯定不会造成新的不符合要求的串，所以只要统计下串中连续的不合法的字符的个数，取最小值就可以了。

    关于造数据：比如第2个样例的k对，也需要考虑一下类似于addcosd这种岔开的，（其实也没事，因为禁止串的两个字符不会同时消除，所以不会产生新的禁止串）

AC代码：

#include <bits/stdc++.h>
#define ll long long
using namespace std;

char s[100000 + 5]; 
int main()
{
	int k,cnt1,cnt2;
	char tmp[15];
	ll ans = 0;
	cin>>s;
	int len = strlen(s);
	cin>>k;
	while(k--) {
		cin>>tmp;
		if(tmp[0] == tmp[1]) {//其实这题不用考虑这个
			for(int i = 0; i<len; i++) {
				while(s[i] == tmp[0] && i<len) ans++,i++;
				ans--;
			}
			continue;
		} 
		cnt1=cnt2=0;
		for(int i = 0; i<len; i++) {
			cnt1=cnt2=0;
			while(s[i] == tmp[0] || s[i] == tmp[1]) {
				if(s[i] == tmp[0]) cnt1++;
				if(s[i] == tmp[1]) cnt2++;
				i++;
			}
			ans += min(cnt1,cnt2);
		}
	}
	printf("%lld\n",ans);
    return 0;
}

或者这样写：（熟悉这种if -> while句式）（很多地方都会用到，比如求约数和【 HDU - 1215 】七夕节（数论，约数和公式））

if(tmp[0] == tmp[1]) {
			for(int i = 0; i<len; i++) {
				if(s[i] == tmp[0] && i<len) {
					i++;
					while(s[i] == tmp[0] && i<len) ans++,i++;
				}
			}
			continue;
		}