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;
void dfs(int u, int p, int **memo, int *lev, int log, vector<int> *g) {
  memo[u][0] = p;
  for (int i = 1; i <= log; i++) memo[u][i] = memo[memo[u][i - 1]][i - 1];
  for (int v : g[u]) {
    if (v != p) {
      lev[v] = lev[u] + 1;
      dfs(v, u, memo, lev, log, g);
    }
  }
}
int lca2(int u, int v, int log, int *lev, int **memo, int **gar) {
  int ans = 1000000;
  if (lev[u] < lev[v]) swap(u, v);
  for (int i = log; i >= 0; i--)
    if ((lev[u] - (1 << i)) >= lev[v]) {
      ans = min(ans, gar[u][i]);
      u = memo[u][i];
    }
  if (u == v) {
    return ans;
  }
  for (int i = log; i >= 0; i--) {
    if (memo[u][i] != memo[v][i]) {
      ans = min(gar[u][i], ans);
      ans = min(gar[v][i], ans);
      u = memo[u][i];
      v = memo[v][i];
    }
  }
  ans = min(gar[u][0], ans);
  ans = min(gar[v][0], ans);
  return ans;
}
void lca1(int u, int v, int log, int *lev, int **memo, int **gar, int &y) {
  if (lev[u] < lev[v]) swap(u, v);
  for (int i = log; i >= 0; i--)
    if ((lev[u] - (1 << i)) >= lev[v]) {
      gar[u][i] = max(y, gar[u][i]);
      u = memo[u][i];
    }
  if (u == v) {
    return;
  }
  for (int i = log; i >= 0; i--) {
    if (memo[u][i] != memo[v][i]) {
      gar[u][i] = max(gar[u][i], y);
      gar[v][i] = max(gar[v][i], y);
      u = memo[u][i];
      v = memo[v][i];
    }
  }
  gar[u][0] = max(gar[u][0], y);
  gar[v][0] = max(gar[v][0], y);
}
int main() {
  int n, m;
  cin >> n;
  vector<int> g[n + 1];
  int log = (int)ceil(log2(n));
  int **memo = new int *[n + 1], **gar = new int *[n + 1];
  for (int i = 0; i < n + 1; i++) {
    memo[i] = new int[log + 1];
    gar[i] = new int[log + 1];
  }
  int *lev = new int[n + 1];
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= log; ++j) {
      memo[i][j] = -1;
      gar[i][j] = 0;
    }
  }
  vector<pair<int, int>> b;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
    b.push_back(make_pair(u, v));
  }
  dfs(1, 1, memo, lev, log, g);
  cin >> m;
  vector<pair<pair<int, int>, int>> q(m);
  for (int i = 0; i < m; ++i) {
    cin >> q[i].first.first >> q[i].first.second >> q[i].second;
    lca1(q[i].first.first, q[i].first.second, log, lev, memo, gar, q[i].second);
  }
  for (int i = log; i; --i) {
    for (int j = 1; j <= n; ++j) {
      gar[j][i - 1] = max(gar[j][i - 1], gar[j][i]);
      gar[memo[j][i - 1]][i - 1] = max(gar[memo[j][i - 1]][i - 1], gar[j][i]);
    }
  }
  for (int i = 1; i <= n; ++i)
    if (gar[i][0] == 0) gar[i][0] = 1000000;
  for (int i = 1; i <= log; ++i) {
    for (int j = 1; j <= n; ++j) {
      gar[j][i] = min(gar[memo[j][i - 1]][i - 1], gar[j][i - 1]);
    }
  }
  for (int i = 0; i < m; ++i) {
    if (lca2(q[i].first.first, q[i].first.second, log, lev, memo, gar) !=
        q[i].second) {
      cout << "-1\n";
      return 0;
    }
  }
  --n;
  for (int i = 0; i < n; ++i)
    cout << lca2(b[i].first, b[i].second, log, lev, memo, gar) << " ";
  return 0;
}