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;
int n, m;
vector<vector<int>> adjlist;
const int nax = 5005;
int color[nax][nax];
int dist[nax];
int trace[nax];
vector<pair<int, int>> edge;
const int INF = 1e7;
struct Query {
  int a, b, val;
  bool operator<(const Query &other) const { return val < other.val; }
};
Query query[nax];
void input() {
  cin >> n;
  adjlist.resize(n + 1);
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    adjlist[u].push_back(v);
    adjlist[v].push_back(u);
    edge.push_back({v, u});
  }
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> query[i].a >> query[i].b >> query[i].val;
  }
  sort(query + 1, query + m + 1);
  for (int i = 0; i < nax; ++i) {
    for (int j = 0; j < nax; ++j) {
      color[i][j] = 0;
    }
  }
}
void dfs(int v, int parent = -1) {
  trace[v] = parent;
  for (auto &c : adjlist[v]) {
    if (c == parent) continue;
    dfs(c, v);
  }
}
void check() {
  for (int i = 1; i <= m; ++i) {
    fill(dist, dist + n + 1, 0);
    fill(trace, trace + n + 1, 0);
    dfs(query[i].a);
    int now = query[i].b;
    int ans = INF;
    while (now != query[i].a) {
      ans = min(ans, color[now][trace[now]]);
      now = trace[now];
    }
    if (ans != query[i].val) {
      cout << -1;
      exit(0);
    }
  }
  for (auto &p : edge) {
    cout << color[p.first][p.second] << ' ';
  }
}
void solve() {
  fill(dist, dist + n + 1, 0);
  fill(trace, trace + n + 1, 0);
  for (int i = 1; i <= m; ++i) {
    fill(dist, dist + n + 1, 0);
    fill(trace, trace + n + 1, 0);
    dfs(query[i].a);
    int now = query[i].b;
    while (now != query[i].a) {
      color[now][trace[now]] = max(color[now][trace[now]], query[i].val);
      color[trace[now]][now] = max(color[trace[now]][now], query[i].val);
      now = trace[now];
    }
  }
  for (auto &p : edge) {
    if (color[p.first][p.second] == 0)
      color[p.first][p.second] = 1000000, color[p.second][p.first] = 1000000;
  }
  check();
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) {
    input();
    solve();
  }
  return 0;
}