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: Wrong Answer)

#include <bits/stdc++.h>
/*#include <iostream>
#include <vector>
#include <math.h>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <string>
#include <cstring>
#include <regex>*/

using namespace std;

using ll = long long;
using dbl = double;
using pii = pair<int, int>;
using pl4 = pair<ll, ll>;
using vi = vector<int>;
using vvi = vector<vi>;
using vs = vector<string>;
using vvs = vector<vs>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vpii = vector<pii>;
using vvpii = vector<vpii>;
using vpl4 = vector<pl4>;
using vvpl4 = vector<vpl4>;
using vd = vector<dbl>;
using vvd = vector<vd>;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pob pop_back()
#define sz size()
#define be begin()
#define en end()
#define asn assign
#define emp empty()
#define ft front()
#define bk back()
#define clr clear()
#define ins insert
#define ers erase
#define res resize

#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define rFOR(i,a,b) for(int i=(b);i>=(a);i--)
#define SORT(a) sort((a).be,(a).en)
#define rSORT(a) sort((a).rbegin(),(a).rend())
#define UNIQUE(a) (a).erase(unique((a).be,(a).en),(a).en)
#define PREVP(a) prev_permutation((a).be,(a).en)
#define NEXTP(a) next_permutation((a).be,(a).en)
#define BINS(a,b) binary_search((a).be,(a).en,(b))
#define LOWB(a,b) (lower_bound((a).be,(a).en,(b))-(a).be)
#define UPB(a,b) (upper_bound((a).be,(a).en,(b))-(a).be)
#define CNT(a,b) count((a).be,(a).en,b)
#define SUM(a) accumulate((a).be,(a).en,0)
#define REV(a) reverse((a).be,(a).en)
#define REGS(a,b) regex_search((a),regex(b))
#define REGM(a,b) regex_match((a),regex(b))
#define yn(a) cout <<((a)?"yes":"no")<<endl;
#define Yn(a) cout <<((a)?"Yes":"No")<<endl;
#define YN(a) cout <<((a)?"YES":"NO")<<endl;
#define say(a) cout <<(a);
#define sal(a) cout <<(a)<<endl;
#define sas cout <<" ";
#define sak cout <<endl;
#define dbg(a) cout <<(#a)<<": "<<(a)<<endl;
#define c2l(a) ((ll)(a-48))
#define a2l(a) ((ll)(a-97))
#define A2l(a) ((ll)(a-65))
#define l2c(a) ((char)(a+48))
#define l2a(a) ((char)(a+97))
#define l2A(a) ((char)(a+65))
#define DigN2(a) ((llabs(a)==0)?(1):((ll)(log2(double(llabs(a))))+1))
#define DigN10(a) ((llabs(a)==0)?(1):((ll)(log10(double(llabs(a))))+1))
#define Dig2(a,b) (((a)>>(b))&1)
#define Dig10(a,b) (ll)(((a)/((ll)(pow(10.0,(double)(b)))))%10)
#define Pow2(a) (1<<(a))
#define Pow10(a) ((ll)(pow(10.0,double(a))))
#define llin(a) ll (a);cin >>(a);
#define stin(a) string (a);cin >>(a);
#define rdn(a,b) ((a)/(b))
#define rou(a,b) ((((double(a)/double(b))-((a)/(b)))<0.5)?((a)/(b)):(((a)/(b))+1))
#define rup(a,b) ((((a)%(b))==0)?((a)/(b)):(((a)/(b))+1))
#define min(a,b) ((a<b)?(a):(b))
#define max(a,b) ((a>b)?(a):(b))
#define powll(a,b) (ll)(pow((double)(a),(double)(b)))
#define Triangle(x1,y1,x2,y2,x3,y3) (((x1)-(x2))*((y1)-(y3))-((x1)-(x3))*((y1)-(y2)))

#define int ll

const ll MOD = 1e9 + 7;
//const ll MOD = 998244353;
//const ll MOD = 9007199254740881;
const ll INF = 1LL << 60;
const string alp = "abcdefghijklmnopqrstuvwxyz";
const string ALP = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

ll gcd(ll a, ll b) { if (b == 0)return a; return gcd(b, a%b); }
ll lcm(ll a, ll b) { return a / gcd(a, b)*b; }

pl4 Bezout(ll a, ll b) {
	if (b != 0) {
		pl4 xy;
		xy = Bezout(b, a%b);
		return mp(xy.se, xy.fi - ((a / b)*xy.se));
	}
	if (b == 0) {
		return mp(1, 0);
	}
}
pl4 Bez(ll a, ll b, ll c) {
	pl4 xy;
	ll x, y, z, gc;
	xy = Bezout(a, b);
	gc = gcd(a, b);
	if (c%gc != 0) return mp(-1, -1);
	x = xy.fi*(c / gc); y = xy.se*(c / gc);
	if (x < 0) z = rup(-x, (b / gc));
	if (x >= 0) z = -x / (b / gc);
	x += z * (b / gc);
	y -= z * (a / gc);
	return mp(x, y);
}

void salv(vll v) {
	say("{");
	FOR(i, 0, v.sz - 1) {
		say(v[i]);
		if (i != v.sz - 1) say(",");
	}
	sal("}")
}

ll DigS10(ll n) {
	ll m = 0;
	FOR(i, 0, DigN10(n) - 1) {
		m += (ll)((llabs(n) % (ll)(pow(10.0, (double)(i + 1)))) / (ll)(pow(10.0, (double)i)));
	}
	return m;
}

ll isP(ll n) {
	if (n <= 1) return 0;
	FOR(i, 2, (ll)sqrt(n)) {
		if (n%i == 0) return 0;
	}
	return 1;
}

vll FactM(1, 1);
vll FactMI(1, 1);

ll PowM(ll a, ll b) {
	ll ans = 1, x = (a%MOD);
	FOR(i, 0, DigN2(b) - 1) {
		if (Dig2(b, i) == 1) ans = (ans*x) % MOD;
		if (i != (DigN2(b) - 1)) x = (x*x) % MOD;
	}
	return ans;
}

void CFactM(ll n) {
	if (FactM.sz <= n) {
		FOR(i, FactM.sz, n) {
			FactM.pb((FactM[i - 1] * (i%MOD)) % MOD);
		}
	}
	return;
}

void CFactMI(ll n) {
	CFactM(n);
	if (FactMI.sz < (n + 1)) FactMI.res(n + 1, -1);
	if (FactMI[n] == -1) FactMI[n] = PowM(FactM[n], MOD - 2);
	rFOR(i, 1, n - 1) {
		if (FactMI[i] != -1) break;
		FactMI[i] = ((FactMI[i + 1] * ((i + 1) % MOD)) % MOD);
	}
	return;
}

ll CombM(ll n, ll k) {
	if ((n < 0) || (k < 0)) return 0;
	if (n < k) return 0;
	if (n + 1 > FactMI.sz) CFactMI(n);
	return ((((FactMI[k] * FactMI[n - k]) % MOD)*FactM[n]) % MOD);
}

signed main() {

	llin(N);
	llin(M);
	vll aa(N + 1, 0);
	aa[1] = 1;
	ll x, y;

	FOR(i, 0, M - 1) {
		cin >> x >> y;
		aa[y] = (aa[x]|aa[y]);
	}
	ll cnt = 0;
	FOR(i, 0, N - 1) {
		cnt += aa[i];
	}
	sal(cnt);

}