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;
std::vector<int> adj[5001];
int edges[5001][5001];
std::vector<pair<int, int>> edges1;
int depth[5001], par[5001];
void dfs(int v, int p, int d) {
  depth[v] = d;
  par[v] = p;
  for (auto c : adj[v])
    if (c != p) dfs(c, v, d + 1);
}
vector<int> getpath(int a, int b) {
  vector<int> ans, path1, path2;
  path1.push_back(a);
  path2.push_back(b);
  while (a != b) {
    if (depth[a] < depth[b])
      path2.push_back(b = par[b]);
    else if (depth[a] > depth[b])
      path1.push_back(a = par[a]);
    else if (depth[a] == depth[b])
      path1.push_back(a = par[a]), path2.push_back(b = par[b]);
  }
  reverse(path2.begin(), path2.end());
  path2.erase(path2.begin());
  path1.insert(path1.end(), path2.begin(), path2.end());
  return ans = path1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
    edges[x][y] = 1000000;
    edges[y][x] = 1000000;
    edges1.push_back({x, y});
  }
  dfs(1, -1, 0);
  std::vector<pair<int, pair<int, int>>> v;
  cin >> m;
  while (m--) {
    int x, y, z;
    cin >> x >> y >> z;
    v.push_back({z, {x, y}});
  }
  sort((v).begin(), (v).end());
  int fl = 0;
  for (auto i : v) {
    int w = i.first;
    int n1 = i.second.first;
    int n2 = i.second.second;
    std::vector<int> p = getpath(n1, n2);
    for (int j = 0; j < p.size() - 1; j++) {
      edges[p[j]][p[j + 1]] = w;
      edges[p[j + 1]][p[j]] = w;
    }
  }
  for (auto i : v) {
    int w = i.first;
    int n1 = i.second.first;
    int n2 = i.second.second;
    std::vector<int> p = getpath(n1, n2);
    int tt = 1e7;
    for (int j = 0; j < p.size() - 1; j++) {
      tt = min(tt, min(edges[p[j]][p[j + 1]], edges[p[j + 1]][p[j]]));
    }
    if (tt != w) {
      fl = 1;
      break;
    }
  }
  if (fl) {
    cout << "-1";
    exit(0);
  }
  for (auto i : edges1) {
    cout << edges[i.first][i.second] << " ";
  }
}