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;
int n, m, second;
vector<int> g[200005];
vector<int> r[200005];
vector<int> cyc[200005];
int visit[200005];
int SCC[200005];
void dfs(int u) {
  visit[u] = 1;
  for (int v : g[u]) {
    if (visit[v]) continue;
    dfs(v);
  }
}
bool color[200005], vis[200005];
stack<int> st;
vector<int> compo[200005];
int mark;
void dfs1(vector<int> g[], int u) {
  color[u] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (!color[v]) dfs1(g, v);
  }
  st.push(u);
  return;
}
void dfs2(vector<int> r[], int u, int mark) {
  compo[mark].push_back(u);
  vis[u] = 1;
  for (int i = 0; i < r[u].size(); i++) {
    int v = r[u][i];
    if (!vis[v]) dfs2(r, v, mark);
  }
  return;
}
void findSCC(vector<int> g[], vector<int> r[], int node) {
  for (int i = 1; i <= node; i++) {
    if (!color[i]) dfs1(g, i);
  }
  while (!st.empty()) {
    int u = st.top();
    st.pop();
    if (!vis[u]) {
      dfs2(r, u, mark);
      mark++;
    }
  }
  return;
}
bool yes(int id) {
  for (int i = 0; i < compo[id].size(); i++) {
    int u = compo[id][i];
    if (cyc[u].size()) return 0;
    if (visit[u]) return 0;
  }
  return 1;
}
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  cin >> n >> m >> second;
  int x[m + 5], y[m + 5];
  for (int i = 0; i < m; i++) {
    cin >> x[i] >> y[i];
    g[x[i]].push_back(y[i]);
    r[y[i]].push_back(x[i]);
  }
  dfs(second);
  findSCC(g, r, n);
  for (int i = 0; i < mark; i++) {
    for (int j = 0; j < compo[i].size(); j++) {
      SCC[compo[i][j]] = i;
    }
  }
  for (int i = 0; i < m; i++) {
    if (SCC[x[i]] != SCC[y[i]]) {
      cyc[y[i]].push_back(x[i]);
    }
  }
  int ans = 0;
  for (int i = 0; i < mark; i++) {
    if (yes(i)) ans++;
  }
  cout << ans << endl;
}
