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 mod = 1000000007;
int INF = 9999999;
long long a, b, c, d, e, f, n, m, k, l;
void createTree(vector<int> &parents, vector<int> &level,
                vector<vector<int> > &mezoblebi) {
  queue<pair<pair<int, int>, int> > q;
  q.push({{1, 1}, 0});
  level[1] = 0;
  while (!q.empty()) {
    int from = q.front().first.first;
    int to = q.front().first.second;
    int lvl = q.front().second;
    q.pop();
    parents[to] = from;
    level[to] = lvl;
    for (int i = 0; i < mezoblebi[to].size(); i++) {
      int newTo = mezoblebi[to][i];
      if (newTo != from) {
        q.push({{to, newTo}, lvl + 1});
      }
    }
  }
}
void makeEdge(vector<vector<int> > &grid, int from, int to, vector<int> &level,
              vector<int> &parent, int value) {
  while (from != to) {
    if (level[from] > level[to]) {
      grid[from][parent[from]] = max(value, grid[from][parent[from]]);
      grid[parent[from]][from] = grid[from][parent[from]];
      from = parent[from];
    } else if (level[from] < level[to]) {
      grid[parent[to]][to] = max(value, grid[parent[to]][to]);
      grid[to][parent[to]] = grid[parent[to]][to];
      to = parent[to];
    } else {
      grid[parent[to]][to] = max(value, grid[parent[to]][to]);
      grid[to][parent[to]] = grid[parent[to]][to];
      grid[from][parent[from]] = max(value, grid[from][parent[from]]);
      grid[parent[from]][from] = grid[from][parent[from]];
      from = parent[from];
      to = parent[to];
    }
  }
}
bool check(vector<vector<int> > &grid, int from, int to, vector<int> &level,
           vector<int> &parent, int value) {
  int minBefore = 99999999;
  while (from != to) {
    if (level[from] > level[to]) {
      minBefore = min(minBefore, grid[from][parent[from]]);
      from = parent[from];
    } else if (level[from] < level[to]) {
      minBefore = min(grid[to][parent[to]], minBefore);
      to = parent[to];
    } else {
      minBefore = min(minBefore, grid[from][parent[from]]);
      minBefore = min(grid[to][parent[to]], minBefore);
      from = parent[from];
      to = parent[to];
    }
  }
  if (value == minBefore) return true;
  return false;
}
int main() {
  int wveroebi;
  cin >> wveroebi;
  vector<pair<int, int> > wiboebi;
  vector<vector<int> > grid(wveroebi + 1);
  vector<vector<int> > mezoblebi(wveroebi + 1);
  for (int i = 0; i <= wveroebi; i++) {
    for (int j = 0; j <= wveroebi; j++) {
      grid[i].push_back(1);
    }
  }
  for (int i = 0; i < wveroebi - 1; i++) {
    int from, to;
    cin >> from >> to;
    wiboebi.push_back({from, to});
    mezoblebi[to].push_back(from);
    mezoblebi[from].push_back(to);
  }
  int q;
  cin >> q;
  vector<pair<pair<int, int>, int> > saboloo(q);
  for (int i = 0; i < q; i++) {
    int from, to, wona;
    cin >> from >> to >> wona;
    saboloo[i] = {{from, to}, wona};
  }
  vector<int> parents(wveroebi + 1, 1);
  vector<int> level(wveroebi + 1, 1);
  createTree(parents, level, mezoblebi);
  for (int i = 0; i < q; i++) {
    int from = saboloo[i].first.first;
    int to = saboloo[i].first.second;
    int value = saboloo[i].second;
    makeEdge(grid, from, to, level, parents, value);
  }
  bool flag = true;
  for (int i = 0; i < q; i++) {
    int from = saboloo[i].first.first;
    int to = saboloo[i].first.second;
    int value = saboloo[i].second;
    if (!check(grid, from, to, level, parents, value)) {
      flag = false;
      break;
    }
  }
  if (!flag) {
    cout << -1;
  } else {
    for (int i = 0; i < wveroebi - 1; i++) {
      int from = wiboebi[i].first;
      int to = wiboebi[i].second;
      cout << grid[from][to] << " ";
    }
  }
  return 0;
}