Problem:
Problem Statement

We have N boxes, numbered 1 through N. At first, box 1 contains one red ball, and each of the other boxes contains one white ball.

Snuke will perform the following M operations, one by one. In the i-th operation, he randomly picks one ball from box x_i, then he puts it into box y_i.

Find the number of boxes that may contain the red ball after all operations are performed.

Constraints

2≤N≤10^5

1≤M≤10^5

1≤x_i,y_i≤N

x_i≠y_i

Just before the i-th operation is performed, box x_i contains at least 1 ball.

Input

The input is given from Standard Input in the following format:

N M
x_1 y_1
:
x_M y_M

Output

Print the number of boxes that may contain the red ball after all operations are performed.

Sample Input 1

3 2
1 2
2 3

Sample Output 1

2

Just after the first operation, box 1 is empty, box 2 contains one red ball and one white ball, and box 3 contains one white ball.

Now, consider the second operation. If Snuke picks the red ball from box 2, the red ball will go into box 3. If he picks the white ball instead, the red ball will stay in box 2.
Thus, the number of boxes that may contain the red ball after all operations, is 2.

Sample Input 2

3 3
1 2
2 3
2 3

Sample Output 2

1

All balls will go into box 3.

Sample Input 3

4 4
1 2
2 3
4 1
3 4

Sample Output 3

3

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, n) for (ll i = 0; i < (ll)(n); ++i)
#define rep2(i, s, n) for (ll i = s; i < (ll)(n); i++)
#define repr(i, n) for (ll i = n; i >= 0; i--)
#define pb push_back
#define COUT(x) cout << (x) << "\n"
#define COUTF(x) cout << setprecision(15) << (x) << "\n"
#define ENDL cout << "\n"
#define DF(x) x.erase(x.begin())
#define ALL(x) x.begin(), x.end()
#define SORT(x) sort(ALL(x))
#define REVERSE(x) reverse(ALL(x))
#define ANS cout << ans << "\n"
#define init() \
  cin.tie(0);  \
  ios::sync_with_stdio(false)
#define LINE cerr << "[debug] line: " << __LINE__ << "\n";
#define debug(x) cerr << "[debug] " << #x << ": " << x << "\n";
#define debugV(v)                       \
  cerr << "[debugV] " << #v << ":";     \
  rep(i, v.size()) cerr << " " << v[i]; \
  cerr << "\n";
using namespace std;
using ll = long long;
using ld = long double;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using mll = map<ll, ll>;
using qll = queue<ll>;
using P = pair<ll, ll>;
using vp = vector<P>;
using vs = vector<string>;

template <typename T>
inline istream& operator>>(istream& i, vector<T>& v) {
  rep(j, v.size()) i >> v[j];
  return i;
}

template <typename T1, typename T2>
inline istream& operator>>(istream& i, pair<T1, T2>& v) {
  return i >> v.first >> v.second;
}

constexpr ll INF = 0x3f3f3f3f3f3f3f3f;
constexpr ld PI = 3.141592653589793238462643383279;
ll get_digit(ll x) {
  return to_string(x).size();
}

ll gcd(ll x, ll y) {
  return y ? gcd(y, x % y) : x;
}

ll lcm(ll a, ll b) {
  return a / gcd(a, b) * b;
}

template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}

template <class T>
bool chmin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}

vector<P> factorize(ll n) {
  vector<P> result;
  for (ll i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      result.pb({i, 0});
      while (n % i == 0) {
        n /= i;
        result.back().second++;
      }
    }
  }
  if (n != 1) {
    result.pb({n, 1});
  }
  return result;
}

vll divisor(ll n) {
  vll ret;
  for (ll i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n) ret.push_back(n / i);
    }
  }
  SORT(ret);
  return (ret);
}

const int mod = 1000000007;
struct mint {
  ll x;
  mint(ll x = 0) : x((x % mod + mod) % mod) {
  }
  mint operator-() const {
    return mint(-x);
  }
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    return mint(*this) += a;
  }
  mint operator-(const mint a) const {
    return mint(*this) -= a;
  }
  mint operator*(const mint a) const {
    return mint(*this) *= a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const {
    return pow(mod - 2);
  }
  mint& operator/=(const mint a) {
    return *this *= a.inv();
  }
  mint operator/(const mint a) const {
    return mint(*this) /= a;
  }
};
istream& operator>>(istream& is, const mint& a) {
  return is >> a.x;
}
ostream& operator<<(ostream& os, const mint& a) {
  return os << a.x;
}

struct combination {
  vector<mint> fact, ifact;
  combination(ll n) : fact(n + 1), ifact(n + 1) {
    assert(n < mod);
    fact[0] = 1;
    for (ll i = 1; i <= n; ++i) {
      fact[i] = fact[i - 1] * i;
    }
    ifact[n] = fact[n].inv();
    for (ll i = n; i >= 1; --i) {
      ifact[i - 1] = ifact[i] * i;
    }
  }
  mint operator()(ll n, ll k) {
    if (k < 0 || k > n) return 0;
    return fact[n] * ifact[k] * ifact[n - k];
  }
};

signed main() {
  init();
  ll N, M;
  cin >> N >> M;
  vll red(N, 1);
  vector<bool> red_c(N, false);
  red[0] = 1;
  red_c[0] = true;

  rep(i, M) {
    ll x, y;
    cin >> x >> y;
    x--;
    y--;

    if (red_c[x]) {
      red[x]--;
      if (red[x] == 0) red_c[x] = false;
      red_c[y] = true;
      red[y]++;
    }
  }

  ll ans = 0;
  rep(i, N) if (red_c[i]) ans++;
  ANS;
  return 0;
}