There are nn cities and mm 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 nn, mm and ss (1≤n≤5000,0≤m≤5000,1≤s≤n1≤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 11 to nn.

The following mm lines contain roads: road ii is given as a pair of cities uiui, vivi (1≤ui,vi≤n1≤ui,vi≤n, ui≠viui≠vi). For each pair of cities (u,v)(u,v), there can be at most one road from uu to vv. Roads in opposite directions between a pair of cities are allowed (i.e. from uu to vv and from vv to uu).

Output

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

For example, you can add roads (6,46,4), (7,97,9), (1,71,7) to make all the cities reachable from s=1s=1.

The second example is illustrated by the following:

In this example, you can add any one of the roads (5,15,1), (5,25,2), (5,35,3), (5,45,4) to make all the cities reachable from s=5s=5.

 

题意：

给定n个节点，M个有向边，和一个节点s。

问最小需要加多少个有向边可以使全部的节点都有到达s节点的路径。


思路：

把除了s节点的其他节点都缩成强连通分量，强连通分量不能到达s节点的，这个分量多加一个边即可到达。

缩成强连通分量的方法可以用dfs+并查集。

枚举每一个边的两边的节点a和b，如果a和b所在的集合（并查集维护出的集合）有一个边联通，那么把这两个节点对应的集合合并（并查集处理合并集合。）

时间复杂度：O（n*m）

 

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=10010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
std::vector<int> v[maxn];
int n,m,rt;
int a,b;
int par[maxn];
void init()
{
    repd(i,1,n)
    {
        par[i]=i;
    }
}
int findpar(int x)
{
    if(par[x]==x)
    {
        return x;
    }else
    {
        return par[x]=findpar(par[x]);
    }
}
int vis[maxn];
bool check(int x,int y)
{
    int res=0;
    if(x==y)
    {
        return 1;
    }
    vis[x]=1;
    for(auto a:v[x])
    {
        if(!vis[a])
        {
            if(check(a,y))
            {
                res=1;
                break;
            }
        }
    }
    return res;
}
int cnt[maxn];
int u[maxn];
int vv[maxn];
int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    gbtb;
    cin>>n>>m>>rt;
    repd(i,1,m)
    {
        cin>>a>>b;
        u[i]=a;
        vv[i]=b;
        v[a].pb(b);
    }
    init();
    repd(i,1,m)
    {
        MS0(vis);
        a=u[i];
        b=vv[i];
        a=findpar(a);
        b=findpar(b);
        if(a!=b&&b!=rt&&check(a,b))
        {
            par[a]=b;
        }
    }
    int ans=0;
    repd(i,1,n)
    {
        if(i!=rt)
        {
            a=findpar(i);
            if(a==i)
            {
                ans++;
            }

        }
    }
    cout<<ans<<endl;
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}