Birdwatching

Kiara studies an odd species of birds that travel in a very peculiar way. Their movements are best explained using the language of graphs: there exists a directed graph $G$ where the nodes are trees, and a bird can only fly from a tree $T_a$ to $T_b$ when $(T_a,T_b)$ is an edge of $G$.
Kiara does not know the real graph $G$ governing the flight of these birds but, in her previous field study, Kiara has collected data from the journey of many birds. Using this, she has devised a graph $P$ explaining how they move. Kiara has spent so much time watching them that she is confident that if a bird can fly directly from $a$ to $b$, then she has witnessed at least one such occurrence. However, it is possible that a bird flew from $a$ to $b$ to $c$ but she only witnessed the stops $a$ and $c$ and then added $(a,c)$ to $P$. It is also possible that a bird flew from $a$ to $b$ to $c$ to $d$ and she only witnessed $a$ and $d$, and added $(a,d)$ to $P$ , etc. To sum up, she knows that $P$ contains all the edges of $G$ and that $P$ might contain some other edges $(a,b)$ for which there is a path from $a$ to $b$ in $G$ (note that $P$ might not contain all such edges).

For her next field study, Kiara has decided to install her base next to a given tree $T$. To be warned of the arrival of birds on $T$, she would also like to install detectors on the trees where the birds can come from (i.e. the trees $T^{\prime }$ such that there is an edge $(T^{\prime },T)$ in $G$ ). As detectors are not cheap, she only wants to install detectors on the trees $T^{\prime }$ for which she is sure that $(T^{\prime },T)$ belongs to $G$.

Kiara is sure that an edge $(a,b)$ belongs to $G$ when $(a,b)$ is an edge of $P$ and all the paths in $P$ starting from $a$ and ending in $b$ use the edge $(a,b)$. Kiara asks you to compute the set $S(T)$ of trees $T^{\prime }$ for which she is sure that $(T^{\prime },T)$ is an edge of $G$.

Input
The input describes the graph $P$. The first line contains three space-separated integers $N$, $M$, and $T$: $N$ is the number of nodes of $P$, $M$ is the number of edges of $P$ and $T$ is the node corresponding to the tree on which Kiara will install her base.

The next $M$ lines describe the edges of the graph $P$. Each contains two space-separated integers $a$ and $b$ $(0\le a,b<Nanda\ne b)$ stating that $(a,b)\in P$ . It is guaranteed that the same pair $(a,b)$ will not appear twice.

Limits

$1\le N,M\le 100000$;
$0\le T<N$.
Output
Your output should describe the set $S(T)$. The first line should contain an integer $L$, which is the number of nodes in $S(T)$, followed by $L$ lines, each containing a different element of $S(T)$. The elements of $S(T)$ should be printed in increasing order, with one element per line.

Examples

input
3 3 2
0 1
0 2
1 2
output
1
1
input
6 8 2
0 1
0 2
1 2
2 0
2 3
3 4
4 2
2 5
output
2
1
4

Note
Sample Explanation 1

The graph corresponding to this example is depicted below. The node $1$ belongs to $S(2)$ because the (only) path from $1$ to $2$ uses $(1,2)$. The node $0$ does not belong to $S(2)$ because the path $0\to 1\to 2$ does not use the edge $(0,2)$.

Sample Explanation 2

The graph corresponding to this example is depicted below. For the same reason as in Sample $1$, the node $0$ does not belong to $S(2)$ while 1 does. The nodes $3$ and $5$ do not belong to $S(2)$ because we do not have edges $(3,2)$ or $(5,2)$. Finally $4$ belongs to $S(2)$ because all paths from $4$ to $2$ use the edge $(4,2)$.
$v1[x]$中存储的元素为$v3[y]=true$且能够到达点$x$的点$y$。
$v2[x]$表示与对于存在边$e_{x,T}$点$x$是否存在其他路径可以到达点$T$。
$v3[x]$表示点$x$是否为存在边$e_{x,T}$点。
将所有有向边反向建图，从所有$v3[x]=true$的点出发$bfs$，如果某个某个点$x$满足$v3[x]=true$并且$v1[x]$中有不是$x$的点，则$v2[x]=true$，最终答案为所有$v2$为$true$的点。
然而如果直接$bfs$会超时，这里需要进行一些优化。
对与要从$x$访问到的点$y$,如果$v1[y]$中有$x$，说明点$y$已经被从$x$出发访问过了；如果$v1[y].size()>1$说明至少有从两个不同点出发的访问到该点，即该点所有出入边都被覆盖过至少一次这两种情况都没必要访问。

#include

#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define sc(a) scahf("%c",&a);
#define ss(a) scanf("%s",a)
#define pi(a) printf("%d\n",a)
#define pl(a) printf("%lld\n",a)
#define pc(a) putchar(a)
#define ms(a) memset(a,0,sizeof(a))
#define repi(i, a, b) for(register int i=a;i<=b;++i)
#define repd(i, a, b) for(register int i=a;i>=b;--i)
#define reps(s) for(register int i=head[s];i;i=Next[i])
#define ll long long
#define ull unsigned long long
#define vi vector
#define pii pair
#define mii unordered_map
#define msi unordered_map
#define lowbit(x) ((x)&(-(x)))
#define ce(i, r) i==r?'\n':' '
#define pb push_back
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define pr(x) cout<<#x<<": "<
using namespace std;

inline int qr() {
    int f = 0, fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')fu = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        f = (f << 3) + (f << 1) + c - 48;
        c = getchar();
    }
    return f * fu;
}

const int N = 1e5 + 10;
int head[N], ver[N << 1], Next[N << 1], tot;
int n, m, st;
vi p;
bool v2[N], v3[N];
unordered_set<int> v1[N];

inline void add(int x, int y) {
    ver[++tot] = y;
    Next[tot] = head[x];
    head[x] = tot;
}

inline void read() {
    n = qr(), m = qr(), st = qr();
    while (m--) {
        int x = qr(), y = qr();
        add(y, x);
    }
}

inline void bfs() {
    queue<pii > q;
    for (auto it:p)q.push({it, it});
    while (!q.empty()) {
        pii x = q.front();
        q.pop();
        reps(x.fi) {
            int y = ver[i];
            if (v1[y].size() >= 2 || v1[y].count(x.se))continue;
            if (v3[y] && x.se != y)v2[y] = true;
            v1[y].insert(x.se), q.push({y, x.se});
        }
    }
}

int main() {
    read();
    reps(st) {
        int y = ver[i];
        p.pb(y), v3[y] = true, v1[y].insert(y), v1[st].insert(y);
    }
    bfs();
    vi ans;
    for (auto it:p)if (!v2[it])ans.pb(it);
    pi(ans.size());
    sort(ans.begin(), ans.end());
    for (auto it:ans)pi(it);
    return 0;
}