Problem: There are n railway stations in Berland. They are connected to each other by n-1 railway sections. The railway network is connected, i.e. can be represented as an undirected tree.

You have a map of that network, so for each railway section you know which stations it connects.

Each of the n-1 sections has some integer value of the scenery beauty. However, these values are not marked on the map and you don't know them. All these values are from 1 to 10^6 inclusive.

You asked m passengers some questions: the j-th one told you three values:

  * his departure station a_j; 
  * his arrival station b_j; 
  * minimum scenery beauty along the path from a_j to b_j (the train is moving along the shortest path from a_j to b_j). 



You are planning to update the map and set some value f_i on each railway section — the scenery beauty. The passengers' answers should be consistent with these values.

Print any valid set of values f_1, f_2, ..., f_{n-1}, which the passengers' answer is consistent with or report that it doesn't exist.

Input

The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of railway stations in Berland.

The next n-1 lines contain descriptions of the railway sections: the i-th section description is two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), where x_i and y_i are the indices of the stations which are connected by the i-th railway section. All the railway sections are bidirected. Each station can be reached from any other station by the railway.

The next line contains a single integer m (1 ≤ m ≤ 5000) — the number of passengers which were asked questions. Then m lines follow, the j-th line contains three integers a_j, b_j and g_j (1 ≤ a_j, b_j ≤ n; a_j ≠ b_j; 1 ≤ g_j ≤ 10^6) — the departure station, the arrival station and the minimum scenery beauty along his path.

Output

If there is no answer then print a single integer -1.

Otherwise, print n-1 integers f_1, f_2, ..., f_{n-1} (1 ≤ f_i ≤ 10^6), where f_i is some valid scenery beauty along the i-th railway section.

If there are multiple answers, you can print any of them.

Examples

Input


4
1 2
3 2
3 4
2
1 2 5
1 3 3


Output


5 3 5


Input


6
1 2
1 6
3 1
1 5
4 1
4
6 1 3
3 4 1
6 5 2
1 2 5


Output


5 3 1 2 1 


Input


6
1 2
1 6
3 1
1 5
4 1
4
6 1 1
3 4 3
6 5 3
1 2 4


Output


-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long md = 1e9 + 7;
const int N = 210000;
const int M = 5100;
struct Edge {
  int v, next;
} edge[N];
int head[N], e;
int un[N], vn[N];
void erase() {
  memset(head, -1, sizeof(head));
  e = 0;
}
void increase(int u, int v) {
  edge[e].v = v;
  edge[e].next = head[u];
  head[u] = e++;
}
struct Path {
  int a, b, g;
} pn[N];
bool cmp(Path a, Path b) { return a.g > b.g; }
int fa[N], w[N], dep[N];
void dfs(int u, int f) {
  fa[u] = f;
  for (int t = head[u]; t != -1; t = edge[t].next) {
    int v = edge[t].v;
    if (v == f) continue;
    dep[v] = dep[u] + 1;
    dfs(v, u);
  }
}
bool update(Path p) {
  int a = p.a, b = p.b, g = p.g;
  bool ok = 0;
  while (a != b) {
    if (dep[a] > dep[b]) {
      if (w[a] <= g) {
        w[a] = g;
        ok = 1;
      }
      a = fa[a];
    } else {
      if (w[b] <= g) {
        w[b] = g;
        ok = 1;
      }
      b = fa[b];
    }
  }
  return ok;
}
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  int i, j, k;
  int u, v;
  int a, b, c, d;
  int z;
  cin >> n;
  erase();
  for (i = 1; i < n; i++) {
    cin >> u >> v;
    increase(u, v);
    increase(v, u);
    un[i] = u, vn[i] = v;
  }
  dep[1] = 1;
  dfs(1, 1);
  cin >> m;
  for (i = 0; i < m; i++) {
    cin >> pn[i].a >> pn[i].b >> pn[i].g;
  }
  sort(pn, pn + m, cmp);
  for (i = 1; i <= n; i++) w[i] = 1;
  bool ok = 1;
  for (i = 0; i < m; i++) {
    if (!update(pn[i])) {
      ok = 0;
      break;
    }
  }
  if (ok) {
    for (i = 1; i < n; i++) {
      if (fa[un[i]] == vn[i])
        cout << w[un[i]] << " ";
      else
        cout << w[vn[i]] << " ";
    }
    cout << endl;
  } else
    cout << -1 << endl;
  return 0;
}