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;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const int maxn = 5000 + 1000;
int n, m, x, y;
vector<vector<pair<int, int> > > g;
vector<vector<pair<int, int> > > p;
vector<int> val;
struct node {
  int a, b, num;
} rev[maxn];
bool cmp(node a, node b) { return a.num < b.num; }
void dfs(int v, int pv, int pidx, vector<pair<int, int> > &p) {
  p[v] = make_pair(pv, pidx);
  for (auto it : g[v]) {
    int to = it.first;
    int idx = it.second;
    if (to == pv) continue;
    dfs(to, v, idx, p);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  cin >> n;
  g = vector<vector<pair<int, int> > >(n);
  val = vector<int>(n, 0);
  for (int i = 0; i < (int)(n - 1); ++i) {
    cin >> x >> y;
    --x;
    --y;
    g[x].push_back({y, i});
    g[y].push_back({x, i});
  }
  p = vector<vector<pair<int, int> > >(n, vector<pair<int, int> >(n));
  for (int i = 0; i < (int)(n); ++i) {
    dfs(i, -1, -1, p[i]);
  }
  cin >> m;
  for (int i = 0; i < (int)(m); ++i) {
    cin >> rev[i].a >> rev[i].b >> rev[i].num;
    --rev[i].a;
    --rev[i].b;
  }
  sort(rev, rev + m, cmp);
  for (int i = 0; i < (int)(m); ++i) {
    int u, v;
    u = rev[i].a;
    v = rev[i].b;
    while (v != u) {
      int pv = p[u][v].first;
      int pidx = p[u][v].second;
      val[pidx] = rev[i].num;
      v = pv;
    }
  }
  int good = (int)1e6;
  for (int i = 0; i < (int)(n - 1); ++i) {
    if (val[i] == 0) val[i] = good;
  }
  for (int i = 0; i < (int)(m); ++i) {
    int u, v;
    u = rev[i].a;
    v = rev[i].b;
    int mx = (int)1e6 + 1000;
    while (v != u) {
      int pv = p[u][v].first;
      int pidx = p[u][v].second;
      mx = min(mx, val[pidx]);
      v = pv;
    }
    if (mx != rev[i].num) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (int i = 0; i < (int)(n - 1); ++i) cout << val[i] << " ";
  cout << '\n';
  return 0;
}