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 N = 1e5 + 10;
int n, m;
vector<int> g[N];
int dfn[N], low[N], sccno[N], clk, scnt;
int s[N], s_top;
void tarjan(int x) {
  dfn[s[++s_top] = x] = low[x] = ++clk;
  for (int y : g[x]) {
    if (!dfn[y])
      tarjan(y), low[x] = min(low[x], low[y]);
    else if (!sccno[y])
      low[x] = min(low[x], dfn[y]);
  }
  if (low[x] == dfn[x])
    for (int u = ++scnt; sccno[u = s[s_top--]] = scnt, u != x;)
      ;
}
int S;
void init() {
  scanf("%d%d%d", &n, &m, &S);
  for (int i = 1; i <= n; ++i) g[i].clear();
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
  }
  clk = scnt = 0;
  memset(sccno, 0, sizeof sccno);
  memset(dfn, 0, sizeof dfn);
  for (int i = 1; i <= n; ++i)
    if (!dfn[i]) tarjan(i);
}
int rd[N];
int main() {
  init();
  for (int i = 1; i <= n; ++i) {
    for (int j : g[i])
      if (sccno[i] != sccno[j]) {
        ++rd[sccno[j]];
      }
  }
  int ans = 0;
  for (int i = 1; i <= scnt; ++i)
    if (!rd[i] && i != sccno[S]) ++ans;
  printf("%d\n", ans);
}
