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 (int i= 0;i<(n);++i)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const long long INF = 1LL << 60;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int main(){

    int n,m;cin>>n>>m;
    vector<int> x(m),y(m);
    vector<int> a(n+1);//個数
    vector<int>red(n+1);//赤がある可能性

    for(int i =1;i<=n;i++){
        a[i]=1;
        red[i]=0;
    }
    red[1]=1;
    rep(i,m){
        cin>>x[i]>>y[i];
    }
    rep(i,m){
        if(a[x[i]]==0)break;
        a[x[i]]-=1;
        a[y[i]]+=1;
        red[y[i]]=red[x[i]];
        if(a[x[i]]==0){
            red[x[i]]=0;
        }


    }
    int count =0;
    for(int i=1;i<=n;i++){
       // cout<<a[i]<<" "<<isred[i]<<endl;
        if(red[i]==1&&a[i]>0)count++;
    }
    cout<<count<<endl;

}