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 = 5010;
struct edge {
  int t, next;
} edges[MAXN];
int head[MAXN], top;
void add(int f, int t) {
  edges[++top].next = head[f];
  edges[top].t = t;
  head[f] = top;
}
int dfn[MAXN], low[MAXN], dfn_cnt, col[MAXN], color;
bool vis[MAXN];
stack<int> stk;
void tarjan(int x) {
  dfn[x] = low[x] = ++dfn_cnt;
  vis[x] = true;
  stk.push(x);
  for (int i = head[x]; i; i = edges[i].next) {
    int t = edges[i].t;
    if (!dfn[t]) {
      tarjan(t);
      low[x] = min(low[x], low[t]);
    } else if (vis[t]) {
      low[x] = min(low[x], dfn[t]);
    }
  }
  if (dfn[x] == low[x]) {
    col[x] = ++color;
    vis[x] = false;
    while (stk.top() != x) {
      col[stk.top()] = color;
      vis[stk.top()] = false;
      stk.pop();
    }
    stk.pop();
  }
}
int indeg[MAXN];
int main() {
  int n, m, s;
  cin >> n >> m >> s;
  int f, t;
  for (int i = 1; i <= m; i++) {
    cin >> f >> t;
    add(f, t);
  }
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) tarjan(i);
  for (int j = 1; j <= n; j++) {
    for (int i = head[j]; i; i = edges[i].next) {
      int t = edges[i].t;
      if (col[t] != col[j]) indeg[col[t]]++;
    }
  }
  int ans = 0;
  for (int i = 1; i <= color; i++) {
    if (!indeg[i] && i != col[s]) ans++;
  }
  cout << ans;
  return 0;
}
