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 fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x = max(l,min(x,r)))
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const v(T)&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>inline ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#define dame { puts("-1"); return 0;}
#define yn {puts("Yes");}else{puts("No");}
const int MX = 200005;
#define RET(ans) {cout<<ans<<endl;return 0;} 

    // 二次元ベクターの基本
    /*
    vector<vector<int>> dp; // 宣言
    dp.resize(n); // 1次元めの要素数決定
    dp[i].push_back(int); // プッシュバック
    rep(i,n){
        sort(dp[i].begin(),dp[i].end()); // 二次元めを昇順ソート
    }
    */

    // 整数スキャン（複数）
    /*
    int x,y,z;
    scanf("%d%d%d",&x,&y,&z);

    // n個の整数のスキャン
    /*
    ll a[n] = {};
    rep(i,n){
        scanf("%lld",&a[i]);
    }
    */

    // 文字列スキャン
    /*
    string s; cin >> s;
    */

    // n個の文字列スキャン
    /*
    vector<string> slist;
    rep(i,n){
        string s; cin >> s;
        slist.push_back(s);
    }
    */

int main() {
    int n,m;
    cin >> n >> m;
    ll x[m] = {};
    ll y[m] = {};
    rep(i,m){
        cin >> x[i] >> y[i];
        x[i]--;
        y[i]--;
    }
    int a[n][2]; // [0]玉の数、[1]赤玉の数（可能性）
    rep(i,n){
        a[i][0] = 1;
        a[i][1] = 0;
    }
    a[0][1] = 1;
    rep(i,m){
        if(a[x[i]][1]==0){
            a[x[i]][0]--;
            a[y[i]][0]++;
        }else{
            if(a[x[i]][0]==1){
                a[x[i]][0]--;
                a[x[i]][1]=0;
                a[y[i]][0]++;
                a[y[i]][1]=1;
            }else{
                a[x[i]][0]--;
                a[y[i]][0]++;
                a[y[i]][1]=1;
            }
        }
    }


    ll ans = 0;
    rep(i,n){
        ans += a[i][1];
    }
    
    cout << ans << endl;
    return 0;
}