#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
#define int ll
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define IN(a, b, x) (a<=x&&x<b)
#define MP make_pair
#define PB push_back
#ifdef int
const ll INF = (1LL<<60);
#else
const int INF = (1LL<<30);
#endif
const double PI = 3.14159265359;
const double EPS = 1e-12;
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

map<PII, VVI> g;
VI x[200010];
map<PII, int> d;
signed main(void)
{
  int n, m, k, s, t;
  cin >> n >> m >> k >> s >> t;
  s--, t--;
  REP(i, m) {
    int a, b, c;
    cin >> a >> b >> c;
    a--, b--, c--;
    // building, floor, cost
    g[{a, b}].PB({a+1, c, 0});
    x[a].PB(b);
    x[a+1].PB(c);
  }
  x[0].PB(s);
  x[n-1].PB(t);
  REP(i, n) {
    sort(ALL(x[i]));
    REP(j, x[i].size()-1) {
      // x[j] -> x[j+1]
      g[{i, x[i][j]}].PB({i, x[i][j+1], x[i][j+1]-x[i][j]});
      g[{i, x[i][j+1]}].PB({i, x[i][j], x[i][j+1]-x[i][j]});
      d[{i, x[i][j]}] = INF;
    }
    d[{i, x[i].back()}] = INF;
  }

  priority_queue<VI, VVI, greater<VI>> que;
  que.push({0, 0, s});
  d[{0, s}] = 0;
  while(que.size()) {
    VI v = que.top(); que.pop();
    int cost = v[0], build = v[1], flr = v[2];
    if(cost > d[{build, flr}]) continue;
    for(auto i: g[{build, flr}]) {
      if(d[{i[0], i[1]}] > d[{build, flr}] + i[2]) {
        d[{i[0], i[1]}] = d[{build, flr}] + i[2];
        que.push({d[{i[0], i[1]}], i[0], i[1]});
      }
    }
  }

  if(d[{n-1, t}] == INF) cout << -1 << endl;
  else cout << d[{n-1, t}] << endl;

  return 0;
}