Problem: G： 検閲により置換 (Censored String)

Problem Statement

You are given the string S, and the set of strings \mathcal P whose size is N. Now we consider applying the following operation to S:

* Choose exactly one character in S, and replace it with '*'.



Let s_i be a i-th character of S, S_{ij} be a consecutive substring in S such that  S = s_i s_{i+1} $\cdots$  s_j, and p_k as the k-th string of P. Your task is to calculate the minimum number of operations for satisfying the following condition.

* For all pairs (S_{ij}, p_k), it is satisfied that S_{ij} \neq p_k.



Input

An input is given in the following format.


S
N
p_1
p_2
$\vdots$
p_N


* In line 1, you are given the string S.
* In line 2, given an integer N. N means the size of P.
* From line 3 to the end, given p_i, the i-th string of P.



Constraints

* 1 \leq |S| \leq 10^5
* 1 \leq N \leq 10^5
* 1 \leq |p_1| + |p_2| +  $\cdots$  + |p_N| \leq 10^5
* If i \neq j, it is guaranteed that p_i \neq p_j
* S, and p_i include lowercase letters only



Output

Print the minimum number of operations in one line.

Sample Input 1


brainfuck
1
fuck


Sample Output 1


1

For example, if you change "brainfuck" into "brainf*ck", you can satisfy the condition. You have to perform operation at least once, and it is the minimum value.

Sample Input 2


aaa
1
a


Sample Output 2


3

You have to operate for all characters.

Sample Input 3


hellshakeyano
3
hell
shake
key


Sample Output 3


2

In this case, for example, if you change "hellshakeyano" into "h*llsha*eyano", you can satisfy the condition. Notice if you replace one character, it might happen that several strings in the set P, which appear as the consecutive substring of S, simultaneously disappear.





Example

Input

brainfuck
1
fuck


Output

1

Solution (in CPP):

Solution:

#if 1
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <stack>
#include <array>
#include <deque>
#include <algorithm>
#include <utility>
#include <cstdint>
#include <functional>
#include <iomanip>
#include <numeric>
#include <assert.h>
#include <bitset>
#include <list>

auto& in = std::cin;
auto& out = std::cout;
#define all(C) std::begin(C), std::end(C)
const double PI = 3.141592653589793238462643383279502884197169399375105820974944;

struct Trie {
	Trie* node[26] = {};//各アルファベットへの子のポインタ
	bool black=false;//この文字列の頂点が黒かどうか
	Trie() :node() {}
	void add(const char* str) {
		if (*str == '\0') {
			black = true;
			return;
		}
		int nxt = *str - 'a';
		if (!node[nxt]) {
			node[nxt] = new Trie;
		}
		node[nxt]->add(str + 1);
	}
	//strを見つけて、最も浅い高さにいるblackな頂点の高さを返す
	//ない場合は-1
	int search(const char* str) {
		if (black)
			return 0;
		if (*str == '\0')
			return -1;
		int nxt = *str - 'a';
		if (!node[nxt]) {
			return -1;
		}
		int res = node[nxt]->search(str + 1);
		if (res == -1)return -1;
		return res + 1;
	}
};

using namespace std;
using ULL = unsigned long long;
const ULL P = 1000000007;
struct RollHash {
	string str;
	vector<ULL>imos;//[0,i)までのハッシュがあるよ
	vector<ULL>pvar;//Pのi乗が入っているよ
	RollHash(string str) :str(str) {
		imos.push_back(0);
		for (int i = 0; i < (int)str.size(); ++i) {
			imos.push_back(imos.back() * P + str[i]);
		}
		pvar.push_back(1);
		for (int i = 0; i <(int)str.size(); ++i) {
			pvar.push_back(pvar.back() * P);
		}
	}
	//[x,y)までのハッシュを返すよ
	ULL getHash(int x, int y) {
		return imos[y] - imos[x] * pvar[y - x];
	}
};

Trie trie;
std::string S;
std::string S_rever;
std::vector<std::string> p_small;
//std::vector<std::string> p_large;
std::vector<RollHash> p_RH;
int32_t N;
int32_t DP_[100010];
int32_t* DP = DP_+2;//[j]はjに*を入れたときの最小値
int main()
{
	using std::endl;
	in.sync_with_stdio(false);
	out.sync_with_stdio(false);
	in.tie(nullptr);
	out.tie(nullptr);

	in >> S >> N; S_rever = S; std::reverse(S_rever.begin(), S_rever.end());
	for (int32_t i = 0; i < N; i++)
	{
		std::string input;
		in >> input;
		if (input.size() <= 317) {
			p_small.emplace_back(std::move(input));
		}
		else {
			//p_large.emplace_back(std::move(input));
			p_RH.emplace_back(std::move(input));
		}
	}


	//小さいほうをとらい木に突っ込む
	for (auto& i : p_small)
	{
		std::reverse(i.begin(), i.end());
		trie.add(i.c_str());
	}

	RollHash S_RH{ S };
	auto iter = S_rever.c_str() + S_rever.size()-1;
	int32_t ok_begin = -1;
	DP[-1] = 0;
	for (int32_t i = 0; i < S.size(); i++)
	{
		//i追加
		DP[i] = DP[ok_begin] + 1;

		//OKを移動
		int res = trie.search(iter);
		if (res == -1) {}
		else {
			ok_begin = std::max(ok_begin, i - res + 1);
		}
		--iter;

		for (auto& P : p_RH)
		{
			int size = P.str.size();
			if (i - size + 1 < 0) { continue; }
			auto bace = S_RH.getHash(i- size+1, i + 1);
			auto other = P.getHash(0, size);
			if (bace == other) {
				ok_begin = std::max(ok_begin, i - size + 1);
			}
		}

	}

	int32_t min_ = 10000000;
	for (int32_t i = ok_begin; i < (int)S.size(); i++)
	{
		min_ = std::min(min_, DP[i]);
	}
	out << min_ << endl;

	return 0;
}
#endif