There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.

What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?

New roads will also be one-way.

Input

The first line of input consists of three integers n, m and s (1 ≤ n ≤ 5000, 0 ≤ m ≤ 5000, 1 ≤ s ≤ n) — the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.

The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).

Output

Print one integer — the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.

Examples

Input

9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1


Output

3


Input

5 4 5
1 2
2 3
3 4
4 1


Output

1

Note

The first example is illustrated by the following:

<image>

For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.

The second example is illustrated by the following:

<image>

In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5.
#include <bits/stdc++.h>
using namespace std;
const int MaxN = 5000 + 5;
const int MaxM = 5000 + 5;
struct Graph {
  int cnte;
  int Head[MaxN], To[MaxM], Next[MaxM];
  Graph() {
    cnte = 0;
    memset(Head, 0, sizeof Head);
    memset(To, 0, sizeof To);
    memset(Next, 0, sizeof Next);
  }
  inline void add_edge(int from, int to) {
    cnte++;
    To[cnte] = to;
    Next[cnte] = Head[from];
    Head[from] = cnte;
  }
};
struct Stack {
  int st[MaxN * 2], sz;
  Stack() {
    memset(st, 0, sizeof st);
    sz = 0;
  }
  inline int size() { return sz; }
  inline bool empty() { return sz == 0; }
  inline int top() { return st[sz]; }
  inline void pop() { sz--; }
  inline void push(int x) { st[++sz] = x; }
};
int N, M, S;
int cntv, V;
Graph G, DAG;
int X[MaxM], Y[MaxM];
int low[MaxN], dfn[MaxN];
int lnk[MaxN];
bool in_stack[MaxN];
Stack st;
bool vis[MaxN];
int ind[MaxN];
void init() {
  scanf("%d %d %d", &N, &M, &S);
  for (int i = 1; i <= M; ++i) {
    int &u = X[i], &v = Y[i];
    scanf("%d %d", &u, &v);
    G.add_edge(u, v);
  }
}
void Tarjan(int u) {
  st.push(u);
  in_stack[u] = true;
  low[u] = dfn[u] = ++cntv;
  for (int i = G.Head[u]; i; i = G.Next[i]) {
    int v = G.To[i];
    if (dfn[v] == 0) {
      Tarjan(v);
      low[u] = min(low[u], low[v]);
    } else if (in_stack[v] == true)
      low[u] = min(low[u], dfn[v]);
  }
  if (low[u] == dfn[u]) {
    V++;
    int v;
    do {
      v = st.top();
      st.pop();
      lnk[v] = V;
      in_stack[v] = false;
    } while (v != u);
  }
}
void dfs(int u) {
  if (vis[u]) return;
  vis[u] = true;
  for (int i = DAG.Head[u]; i; i = DAG.Next[i]) dfs(DAG.To[i]);
}
void solve() {
  for (int i = 1; i <= N; ++i)
    if (dfn[i] == 0) Tarjan(i);
  for (int i = 1; i <= M; ++i) {
    int u = X[i], v = Y[i];
    if (lnk[u] == lnk[v]) continue;
    DAG.add_edge(lnk[u], lnk[v]);
    ind[lnk[v]]++;
  }
  dfs(lnk[S]);
  int ans = 0;
  for (int i = 1; i <= V; ++i)
    if (ind[i] == 0 && vis[i] == false) dfs(i), ans++;
  for (int i = 1; i <= V; ++i)
    if (vis[i] == false) dfs(i), ans++;
  cout << ans << endl;
}
int main() {
  init();
  solve();
  return 0;
}
