#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define pb push_back
#define dbg(x) cerr << #x << " " << x << "\n"
/**
-> We use a simple tree dp:
dp[node][size][0/1] = the minimum time it takes to visit size nodes in the subtree of node if we start in node
0 -> we don't visit all
1 -> we visit all
**/
const int N = 10000;
vector <pair <int, int>> gr[1 + N];
int dp[1 + N][1 + N][2];
int sz[1 + N];
void _ (int &a, int b) {
if (a > b)
a = b;
}
void dfs (int node, int par) {
sz[node] = 1;
dp[node][1][0] = dp[node][1][1] = 0;
for (pair <int, int> edge : gr[node]) {
int son = edge.first;
int cost = edge.second;
if (par != son) {
dfs (son, node);
for (int cur = sz[node]; cur >= 0; cur--)
for (int add = sz[son]; add > 0; add--) {
_ (dp[node][cur + add][0], dp[node][cur][0] + dp[son][add][0] + 2 * cost);
_ (dp[node][cur + add][1], dp[node][cur][0] + dp[son][add][1] + cost);
_ (dp[node][cur + add][1], dp[node][cur][1] + dp[son][add][0] + 2 * cost);
}
sz[node] += sz[son];
}
}
}
int main() {
ios::sync_with_stdio (false);
cin.tie (0); cout.tie (0);
int n, k, x;
cin >> n >> k >> x;
for (int i = 1; i < n; i++) {
int x, y, c;
cin >> x >> y >> c;
gr[x].pb ({y, c});
gr[y].pb ({x, c});
}
for (int node = 1; node <= n; node++)
for (int i = 1; i <= n; i++)
dp[node][i][0] = dp[node][i][1] = (1 << 29);
dfs (x, 0);
cout << dp[x][k][1] << "\n";
return 0;
}
