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.
n,m,s=map(int,raw_input().split())
l=[]
from copy import *
for _ in range(0,6000):
  l.append([])
for i in range(0,m):
 # a=i+1
  #b=i+2
  a,b=map(int,raw_input().split())
  l[a].append(b)
  #l[b].append(a)
vis=[-1]*(5004)
#print l
ans=[]
def intersection(lst1, lst2):

    temp = set(lst2)
    lst3 = [value for value in lst1 if value in temp]
    return lst3
def dfs(i,ans):
  st=[]
  st.append(i)
  while len(st)>0:
    val=st.pop()
    if vis[val]==-1:
      ans.append(val)
      vis[val]=0
      for j in l[val]:
        if vis[j]==-1:
          st.append(j)
  return ans
dfs(s,[])
ans2=[]

for i in range(1,n+1):
  if vis[i]==-1:
    ans2.append(i)
fn=[[0,0]]*5004
#print ans2
for i in range(0,len(ans2)):
  val=ans2[i]
  vis=[-1]*5004
  te=dfs(val,[])
  #print te,val
  c=0
  
  for i in range(0,len(ans2)):
    if vis[ans2[i]]!=-1:
      c=c+1 
    
  #fe=list(set(te).intersection(set(ans2)))
  fn[val]=[c,val]
#print fn
fn.sort(reverse=True)
put=[-1]*5004
ans=0
#print fn[0:5]

for i in range(0,len(fn)):
  if len(fn[i])>1 and fn[i][0]!=0  and put[fn[i][1]]==-1 :
    val=fn[i][1]
    put[val]=0
    ans=ans+1
    vis=[-1]*5004
    k=dfs(val,[])
    #print k,val
    for j in range(0,len(k)):
      put[k[j]]=0
  
print ans