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 = 1e5 + 4;
bool erased[N];
struct Edge {
  int v;
  long long cost;
  int i;
  int zeros;
  int p;
  bool operator<(const Edge &e) const {
    if (cost != e.cost) return cost > e.cost;
    return zeros > e.zeros;
  }
};
vector<Edge> adj[N];
Edge parent[N];
long long dijkstra(int source, int destination) {
  long long dist[N];
  memset(parent, -1, sizeof parent);
  memset(dist, 0x3f, sizeof dist);
  priority_queue<Edge> q;
  q.push({source, 0, -1, 0});
  while (!q.empty()) {
    auto cur = q.top();
    q.pop();
    if (dist[cur.v] < 1e18) continue;
    dist[cur.v] = cur.cost;
    parent[cur.v] = {cur.p, 0, cur.i, cur.zeros};
    if (cur.v == destination) break;
    for (auto child : adj[cur.v])
      if (dist[child.v] > 1e18)
        q.push({child.v, cur.cost + child.cost, child.i,
                cur.zeros + erased[child.i], cur.v});
  }
  return dist[destination];
}
int u[N], v[N], c[N];
int main() {
  int n, m, l, s, t;
  scanf("%d %d %d %d %d", &n, &m, &l, &s, &t);
  for (int i = 0; i < m; ++i) {
    scanf("%d %d %d", u + i, v + i, c + i);
    erased[i] = c[i] == 0;
    if (!c[i]) c[i] = 1;
    adj[u[i]].push_back({v[i], c[i], i});
    adj[v[i]].push_back({u[i], c[i], i});
  }
  long long dis = dijkstra(s, t);
  if (dis > l || parent[t].zeros == 0 && dis < l) {
    puts("NO");
    return 0;
  }
  int node = t;
  bool first = false;
  bool leave[N] = {};
  while (node != s) {
    if (erased[parent[node].i]) {
      leave[parent[node].i] = true;
      if (!first++)
        c[parent[node].i] = max(1, c[parent[node].i] + l - (int)dis);
    }
    node = parent[node].v;
  }
  for (int i = 0; i < m; ++i)
    if (erased[i] && !leave[i]) c[i] = 2e9;
  for (int i = 0; i < n; ++i)
    for (auto &child : adj[i]) child.cost = c[child.i];
  if (dijkstra(s, t) != l) {
    puts("NO");
  } else {
    puts("YES");
    for (int i = 0; i < m; ++i) printf("%d %d %d\n", u[i], v[i], c[i]);
  }
  return 0;
}
