Problem:
Score : 300 points

Problem Statement

You drew lottery N times. In the i-th draw, you got an item of the kind represented by a string S_i.

How many kinds of items did you get?

Constraints

1 \leq N \leq 2\times 10^5

S_i consists of lowercase English letters and has a length between 1 and 10 (inclusive).

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

Print the number of kinds of items you got.

Sample Input 1

3
apple
orange
apple

Sample Output 1

2

You got two kinds of items: apple and orange.

Sample Input 2

5
grape
grape
grape
grape
grape

Sample Output 2

1

Sample Input 3

4
aaaa
a
aaa
aa

Sample Output 3

4

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

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

using i64 = int64_t;

#define rep(i, x, y) for (i64 i = i64(x), i##_max_for_repmacro = i64(y); i < i##_max_for_repmacro; ++i)
#define debug(x) #x << "=" << (x)

#ifdef DEBUG
#define _GLIBCXX_DEBUG
#define print(x) std::cerr << debug(x) << " (L:" << __LINE__ << ")" << std::endl
#else
#define print(x)
#endif

template <i64 p>
class fp {
    public:
    i64 x;
    fp() : x(0) {}
    fp(i64 x_) : x((x_ % p + p) % p) {}
    fp operator+() const { return fp(x); }
    fp operator-() const { return fp(-x); }
    fp& operator+=(const fp& y) {
        x += y.x;
        if (x >= p) x -= p;
        return *this;
    }
    fp& operator-=(const fp& y) { return *this += -y; }
    fp& operator*=(const fp& y) {
        x = x * y.x % p;
        return *this;
    }
    fp& operator/=(const fp& y) { return *this *= fp(inverse(y.x)); }
    fp operator+(const fp& y) const { return fp(x) += y; }
    fp operator-(const fp& y) const { return fp(x) -= y; }
    fp operator*(const fp& y) const { return fp(x) *= y; }
    fp operator/(const fp& y) const { return fp(x) /= y; }
    bool operator==(const fp& y) const { return x == y.x; }
    bool operator!=(const fp& y) const { return !(*this == y); }
    i64 extgcd(i64 a, i64 b, i64& x, i64& y) {
        i64 d = a;
        if (b != 0) {
            d = extgcd(b, a % b, y, x);
            y -= (a / b) * x;
        } else {
            x = 1;
            y = 0;
        }
        return d;
    }
    i64 inverse(i64 a) {
        i64 x, y;
        extgcd(a, p, x, y);
        return (x % p + p) % p;
    }
};

template <i64 p>
i64 abs(const fp<p>& x) { return x.x; }

template <i64 p>
istream& operator>>(istream& is, fp<p>& x) {
    is >> x.x;
    return is;
}

template <i64 p>
ostream& operator<<(ostream& os, const fp<p>& x) {
    os << x.x;
    return os;
}

template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
    os << "(" << p.first << ", " << p.second << ")";
    return os;
}

template <typename T>
ostream& operator<<(ostream& os, const vector<T>& vec) {
    os << "[";
    for (const auto& v : vec) {
        os << v << ",";
    }
    os << "]";
    return os;
}

template <typename T>
bool chmin(T& a, const T& b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template <typename T>
bool chmax(T& a, const T& b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template <typename A, typename T, size_t size>
void fill(A (&ary)[size], const T& val) {
    fill((T*)ary, (T*)(ary + size), val);
}

constexpr int inf = 1.01e9;
constexpr i64 inf64 = 4.01e18;
constexpr long double eps = 1e-9;

// double(64bit浮動小数)のn分探索のループ回数の上限(2分探索なら50でも十分かもしれない). long double(80ビットの x87 浮動小数点型?)だと, 2分探索であってもこれだと足りないケースがある気がするので, もうちょっと余裕を持たせた方が良さそう.
constexpr i64 max_loop = 100;

void solve() {
    //constexpr i64 mod = 1'000'000'007;
    i64 N;
    cin >> N;

    set<string> st;
    rep(i,0,N){
        string S;
        cin >> S;
        st.insert(S);
    }

    cout << st.size() << endl;
}

int main() {
    std::cin.tie(0);
    std::ios::sync_with_stdio(false);
    cout.setf(ios::fixed);
    cout.precision(16);
    solve();
    return 0;
}