#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
//对pair的一种有效快捷的利用
#define fi first
#define se second
#define mp make_pair
using namespace std;

typedef long long ll;
const int maxn = 2222222;
ll dis[maxn];
int vis[maxn], ord[maxn], deep[maxn], sz[maxn];
int p[maxn][22];
//dij时用的堆
set <pair<ll, int> > hs;
//利用pair简化邻接表
vector < pair<int, ll>> e[maxn];
const ll inf = 1ll<<60;
void Dijk(int S, int n)
{
    for(int i = 1; i <= n; i++) dis[i] = inf, vis[i] = 0;
    dis[S] = 0;
    for(int i = 1; i <= n; i++) hs.insert(mp(dis[i], i));
    for(int i = 0; i < n; i++)
    {
        int u = (hs.begin())->se; hs.erase(hs.begin());
        vis[u] = 1; ord[i] = u; //求最短路时顺便得到拓扑序orz
        for(int j = 0; j < e[u].size(); j++)
        {
            int v = e[u][j].fi;
            if(dis[v] > dis[u] + e[u][j].se)
            {
                hs.erase(mp(dis[v], v));
                dis[v] = dis[u] + e[u][j].se;
                hs.insert(mp(dis[v], v));
            }
        }
    }
}

int lca(int u, int v)   //二进制倍增求LCA
{
    if(deep[u] > deep[v]) swap(u, v);
    for(int i = 20; i >= 0; i--) if(deep[p[v][i]] >= deep[u]) v = p[v][i];
    if(u == v) return u;
    for(int i = 20; i >= 0; i--) if(p[v][i] != p[u][i]) u = p[u][i], v = p[v][i];
    return p[u][0];
}
int n, m, s, u, v, w;
int main()
{
    //freopen("a.txt", "r", stdin);
    cin>>n>>m>>s;
    for(int i = 1; i <= m; i++)
    {
        cin>>u>>v>>w;
        e[u].push_back(mp(v, w));
        e[v].push_back(mp(u, w));
    }
    Dijk(s, n);
    p[s][0] = 0; deep[s] = 1;
    //构建最短路图的过程并建立支配树
    for(int i = 1; i <= n; i++)
    {
        int d = -1, u = ord[i];
        for(auto p : e[u])
        {
            if(dis[p.fi] + p.se == dis[u])
            {
                if(d == -1) d = p.fi;
                else d = lca(d, p.fi);
            }
        }
        p[u][0] = d; deep[u] = deep[d]+1;
        for(int j = 1; j < 21; j++) p[u][j] = p[p[u][j-1]][j-1];    //动态更新公共祖先
    }
    for(int i = 1; i <= n; i++) sz[i] = 1;
    int ret = 0;
    for(int i = n-1; i >= 1; i--)       //按照拓扑序dp求最大值
    {
        u = ord[i];
        sz[p[u][0]] += sz[u];
        if(dis[u] <= (1ll<<50)) ret = max(ret, sz[u]);
    }
    cout<<ret<<endl;
}