7 9 999999999 0 3
0 1 0
1 2 0
2 3 0
0 4 1
4 1 1
1 5 499999999
5 2 499999999
2 6 1
6 3 1
 7 9 320 0 3
0 1 0
1 2 0
2 3 0
0 4 1
4 1 1
1 5 100
5 2 100
2 6 59
6 3 61
 4 5 7 0 3
0 1 0
1 2 3
2 3 0
0 2 5
1 3 5
 8 9 10 1 0
1 2 1
2 4 1
1 3 0
3 4 0
4 5 0
5 6 1
6 0 1
5 7 0
7 0 0
 4 4 14 1 3
1 3 13
2 3 0
2 0 0
1 0 12
 5 6 1000000000 0 4
0 1 1
2 0 2
3 0 3
4 1 0
4 2 0
3 4 0
 4 5 10 1 2
0 1 3
1 2 0
1 3 4
2 3 4
2 0 6
 4 4 2 1 3
1 3 13
2 3 0
2 0 0
1 0 0
 5 5 2 0 2
0 1 1
1 2 1
0 4 0
4 3 0
3 2 0
 5 5 3 0 2
0 1 1
1 2 1
0 4 0
4 3 0
3 2 0
 4 4 8 1 3
1 3 13
2 3 0
2 0 0
1 0 6
 1000 1 1000000000 998 0
0 999 0
 5 5 1 0 2
0 1 1
1 2 1
0 4 0
4 3 0
3 2 0
 100 1 123456 99 0
0 99 123456
 4 4 13 1 3
1 3 13
2 3 0
2 0 0
1 0 12
 7 9 319 0 3
0 1 0
1 2 0
2 3 0
0 4 1
4 1 1
1 5 100
5 2 100
2 6 59
6 3 61
 1000 1 5 999 0
0 999 0

YES
0 1 1
1 2 999999996
2 3 999999997
0 4 1
4 1 1
1 5 499999999
5 2 499999999
2 6 1
6 3 1
 YES
0 1 1
1 2 199
2 3 318
0 4 1
4 1 1
1 5 100
5 2 100
2 6 59
6 3 61
 YES
0 1 3
1 2 3
2 3 2
0 2 5
1 3 5
 YES
1 2 1
2 4 1
1 3 1
3 4 1
4 5 6
5 6 1
6 0 1
5 7 1
7 0 1
 NO YES
0 1 1
2 0 2
3 0 3
4 1 999999999
4 2 999999998
3 4 999999997
 NO NO YES
0 1 1
1 2 1
0 4 1
4 3 1
3 2 1
 NO YES
1 3 13
2 3 1
2 0 1
1 0 6
 NO NO YES
0 99 123456
 YES
1 3 13
2 3 1
2 0 1
1 0 12
 YES
0 1 1
1 2 198
2 3 317
0 4 1
4 1 1
1 5 100
5 2 100
2 6 59
6 3 61
 YES
0 999 5

#include <bits/stdc++.h>
using namespace std;
const int N = 1010, M = 1e4 + 10;
int n, m, L, s, t, w[M];
vector<pair<int, int>> adj[N];
long long to[N], from[N];
void dijkstra(int u, long long dist[], bool all) {
  for (int i = 0; i < n; i++) {
    dist[i] = 1e18;
  }
  dist[u] = 0;
  set<pair<long long, int>> s;
  s.insert({0, u});
  while (!s.empty()) {
    auto tmp = *s.begin();
    s.erase(s.begin());
    long long d = tmp.first;
    int u = tmp.second;
    if (dist[u] == d) {
      for (auto tmp : adj[u]) {
        int v = tmp.first, cost = w[tmp.second];
        if (!cost && !all) continue;
        if (!cost) cost = 1;
        if (dist[u] + cost < dist[v]) {
          dist[v] = dist[u] + cost;
          s.insert({dist[v], v});
        }
      }
    }
  }
}
void fail() {
  printf("NO\n");
  exit(0);
}
int main() {
  scanf("%d %d %d %d %d", &n, &m, &L, &s, &t);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d %d %d", &u, &v, &w[i]);
    adj[u].push_back({v, i});
    adj[v].push_back({u, i});
  }
  dijkstra(s, to, false);
  if (to[t] < L) fail();
  if (to[t] > L) {
    dijkstra(s, to, true);
    dijkstra(t, from, true);
    if (to[t] > L) fail();
    int u = s;
    bool first = true;
    while (u != t) {
      for (auto tmp : adj[u]) {
        int v = tmp.first, &cost = w[tmp.second];
        if (to[u] + max(cost, 1) + from[v] == to[t]) {
          if (!cost) {
            if (first) {
              cost = L - to[t] + 1;
              first = false;
            } else {
              cost = 1;
            }
          }
          u = v;
          break;
        }
      }
    }
  }
  printf("YES\n");
  for (int u = 0; u < n; u++) {
    for (auto tmp : adj[u]) {
      int v = tmp.first, cost = w[tmp.second];
      if (v < u) continue;
      long long final_cost = (cost ? cost : 1e18);
      printf("%d %d %lld\n", u, v, final_cost);
    }
  }
  return 0;
}
