#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 5e4 + 30;
const int M = 5e5 + 3;

struct node {
    int x, y, val, opt;
}e[M];
int n, m, ans, k, num, fa[N], s;

bool cmp(node a, node b) {
    if(a.val == b.val) return a.opt < b.opt;
    return a.val < b.val;
}

int getfa(int x) {
    if(fa[x] == x) return x;
    return fa[x] = getfa(fa[x]);
}

bool check(int mid) {
    num = ans = 0;
    int tot = 0, x, y, fx, fy;
    for(int i = 1; i <= m; ++i)
        if(e[i].opt == 0) e[i].val -= mid;
    sort(e + 1, e + 1 + m, cmp);
    for(int i = 1; i <= n; ++i) fa[i] = i;
    for(int i = 1; i <= m; ++i) {
        x = e[i].x; y = e[i].y;
        fx = getfa(x); fy = getfa(y);
        if(fx == fy) continue;
        ans += e[i].val; fa[fx] = fy;
        if(e[i].opt == 0) num++;
        tot++;
        if(tot == n - 1) break;
    }
    for(int i = 1; i <= m; ++i)
        if(e[i].opt == 0) e[i].val += mid;
    return num >= k;
}

bool Check(int mid) {
    num = ans = 0;
    int tot = 0, x, y, fx, fy;
    for(int i = 1; i <= m; ++i)
        if(e[i].opt == 0) e[i].val -= mid;
    sort(e + 1, e + 1 + m, cmp);
    for(int i = 1; i <= n; ++i) fa[i] = i;
    for(int i = 1; i <= m; ++i) {
        x = e[i].x; y = e[i].y;
        fx = getfa(x); fy = getfa(y);
        if(fx == fy) continue;
        ans += e[i].val; fa[fx] = fy;
        if(e[i].opt == 0) num++;
        tot++;
        if(tot == n - 1) break;
    }
    for(int i = 1; i <= m; ++i)
        if(e[i].opt == 0) e[i].val += mid;
    return num > k;
}

signed main() {
    int x, y, z, opt;
    scanf("%lld%lld%lld%lld", &n, &m, &k, &s);
    for(int i = 1; i <= m; ++i) {
        scanf("%lld%lld%lld", &x, &y, &z);
        if (x == s || y == s) opt = 0;
        else opt = 1;
        e[i] = {x, y, z, opt};
    }
    int l = -2e9, r = 2e9, mid;
    if (Check(l)) return puts("nonnondayo"), 0;
    if (!check(r)) return puts("nonnondayo"), 0;
    while(l + 1 < r) {
        mid = (l + r) >> 1;
        if(check(mid)) r = mid;
        else l = mid;
    }
    if (check(l)) printf("%lld\n", ans + k * l);
    else check(r), printf("%lld\n", ans + k * r);
	return 0;
}