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 ll long long
#define REV(v) reverse(v.begin(), v.end());
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, start, stop) for (int i = start; i < stop; i++)
#define FORR(i, start, stop) for (int i = start; i > stop; i--)
#define SORT(v, n) sort(v, v + n);
#define SORTR(v, n) sort(v, v + n,greater<ll>());
#define VSORT(v) sort(v.begin(), v.end());
#define VSORTR(v) sort(v.begin(), v.end(),greater<ll>());
#define REMOVE(v,n) remove(vector<ll> v,v+v.size(),ll n)
#define ERASE(v,n) v.erase(v.begin()+n);
#define ull unsigned long long
#define pb(a) push_back(a)
#define INF 999999999
#define V(v,i,j) vector(v.begin() + i, v.begin() + j)
#define INSERT(va,vb) va.insert(va.end(), vb.begin(), vb.end())
using namespace std;
typedef vector<ll> vll;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;
int dy[] = {0, 0, 1, -1};
int dx[] = {1, -1, 0, 0};
const int MOD = 1000000007;
void chmin(int &a, int b) { if (a > b) a = b; }

//vector< vector<int> > v (size1, vector<int>(size2) );

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    ll N,M;
    cin>>N>>M;
    vll ball(N,1); //ボールの個数を格納
    vector<bool> red(N,false); //赤いボールが入っている可能性のある箱をture
    red[0]=true;

    ll x,y;
    FOR(i,0,M){
        cin>>x>>y;x--;y--;
        if(red[x]) red[y]=true;
        if(ball[x]==1) red[x]=false;
        ball[x]--;ball[y]++;
    }

    //集計
    ll ans=0;
    FOR(i,0,N)
        if(red[i])ans++;

    cout<<ans<<endl;

    return 0;
}