/*
我对题目2的思路：
根据每个字母的约束条件，找到满足条件的最长子串，答案就是长度减去这个最长子串的长度。
*/
#include <cstdio>  
#include <string.h>   
#include <algorithm>  
using namespace std;

const int MAX = 100010;
char str[MAX];
int invalid[30][30], dp[26];

int main()
{
	int n, m;
	while (scanf("%d", &n) != EOF){
		scanf("%s", str);
		scanf("%d", &m);

		memset(invalid, 0, sizeof(invalid));
		memset(dp, 0, sizeof(dp));

		for (int i = 0; i<m; i++){
			char ch1, ch2;
			getchar();
			scanf("%c%c", &ch1, &ch2);
			invalid[ch1 - 'a'][ch2 - 'a'] = invalid[ch2 - 'a'][ch1 - 'a'] = 1;
		}

		for (int i = 0; i<n; i++){
			char a = str[i];
			int tmp = 1;
			for (int j = 0; j<26; j++){
				if (invalid[a - 'a'][j]) continue;
				tmp = max(tmp, dp[j] + 1);
			}
			//printf("%d ", tmp);
			dp[a - 'a'] = tmp;
		}
		
		sort(dp, dp + 26);
		printf("%d\n", n - dp[25]);
	}
}

/*
我对题目3的思路：
用队列模拟注册流程，需要判断到达时间，决定办理先后顺序
*/
#include <iostream>
#include <vector>
#include <queue>
using namespace std;

const int N = 10010;
const int INF = 0x3f3f3f3f;

int n, m, k;
struct node
{
	int now;   //arrive time Ti
	int id;    //student number Si;
	int index;
	int len;   //visit number of Pi offices
	int v;     //index
	vector<int> s;  //Oij offices
	vector<int> t;  //Wij processing time
	node(){
		now = id = index = len = v = 0;
	};
};

node *pNode[N];
int res[N], beg[105];

struct mycmp
{
	bool operator()(node *a, node *b){
		if (a->now == b->now)
			return a->id > b->id;
		return a->now>b->now;
	}
};

priority_queue<node*, vector<node*>, mycmp> q;

int main()
{
	scanf("%d %d %d", &n, &m, &k);
	for (int i = 0; i<n; i++){
		int num; //num is Pi offices
		pNode[i] = new node();
		scanf("%d %d %d", &pNode[i]->id, &pNode[i]->now, &num);
		pNode[i]->v = i;
		pNode[i]->len = num;

		int a, b; //Oij and Wij
		for (int j = 0; j<num; j++){
			scanf("%d %d", &a, &b);
			pNode[i]->s.push_back(a);
			pNode[i]->t.push_back(b);
		}

		pNode[i]->index = 0;//start from first office
		pNode[i]->now += k;//from gate to first office
		q.push(pNode[i]);
	}

	while (!q.empty()){
		node *tmp = q.top();
		q.pop();
		int b = max(tmp->now, beg[tmp->s[tmp->index]]);
		tmp->now = beg[tmp->s[tmp->index]] = b + tmp->t[tmp->index];
		if (tmp->index == tmp->len - 1){
			//end of registration
			res[tmp->v] = tmp->now;
		}
		else{
			tmp->now += k;//from one office to another
			tmp->index++;
			q.push(tmp);
		}
	}

	for (int i = 0; i<n; i++)
		printf("%d\n", res[i]);
	return 0;
}