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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;
typedef long long ll;
template<class T>inline bool chmax(T &a, const T &b){if(a < b){a = b; return 1;}return 0;}
template<class T>inline bool chmin(T &a, const T &b){if(a > b){a = b; return 1;}return 0;}
int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  // AGC002B - Box and Ball
  // 1からNまでの番号が振られたN個の箱
  // 1番目の箱には赤いボール、2からN番目の箱には白いボールが1個ずつ入っている
  // M回操作、x番目の箱からボールを1個選びy番目の箱に移動
  // 赤いボールが入っている可能性のある箱の数を求める
  // 基本はbool型の配列を準備して、論理和を求める
  // ただし、箱が空っぽになった場合は元の箱をfalseにしなければならない
  // 個数管理の配列も準備する
  int N, M;
  cin >> N >> M;
  vector<int> x(M);
  vector<int> y(M);
  REP(i, M){
    cin >> x.at(i) >> y.at(i);
  }
  // bool型の配列（赤いボールが入っている可能性があればtrue
  vector<bool> red(N+1, false);
  red[1] = true;
  // 個数管理の配列
  vector<int> cnt(N+1, 1);
  REP(i, M){
    red[y.at(i)] = red[x.at(i)] || red[y.at(i)];
    cnt[y.at(i)]++;
    cnt[x.at(i)]--;
    if(cnt[x.at(i)] == 0) red[x.at(i)] = false;
  }
  int ans = 0;
  REP(i, N+1){
    ans += red.at(i);
  }
  cout << ans << endl;
}