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;
const int nmax = 5e3 + 5;
vector<int> l[nmax];
vector<pair<int, int>> edges;
vector<pair<int, pair<int, int>>> q;
int tata[nmax], d[nmax], ans[nmax], sol[nmax];
void dfs(int nod) {
  for (auto x : l[nod])
    if (x != tata[nod]) {
      tata[x] = nod;
      d[x] = d[nod] + 1;
      dfs(x);
    }
}
void update(int x, int y, int g) {
  if (d[x] < d[y]) swap(x, y);
  while (d[x] > d[y]) {
    ans[x] = max(ans[x], g);
    x = tata[x];
  }
  while (x != y) {
    ans[x] = max(ans[x], g);
    ans[y] = max(ans[y], g);
    x = tata[x], y = tata[y];
  }
}
bool check() {
  for (auto z : q) {
    int mn = INT_MAX;
    int x = z.second.first, y = z.second.second;
    if (d[x] < d[y]) swap(x, y);
    while (d[x] > d[y]) {
      mn = min(mn, ans[x]);
      x = tata[x];
    }
    while (x != y) {
      mn = min(mn, min(ans[x], ans[y]));
      x = tata[x], y = tata[y];
    }
    if (mn != z.first) return 0;
  }
  return 1;
}
int main() {
  int n, m, x, y, g;
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    l[x].push_back(y);
    l[y].push_back(x);
    edges.push_back({x, y});
  }
  dfs(1);
  cin >> m;
  while (m--) {
    cin >> x >> y >> g;
    q.push_back({g, {x, y}});
    update(x, y, g);
  }
  for (int i = 1; i <= n; i++)
    if (ans[i] == 0) ans[i] = 1e6;
  if (check() == 1) {
    int ind = 0;
    for (auto x : edges) {
      ind++;
      if (x.first != tata[x.second]) swap(x.first, x.second);
      sol[ind] = ans[x.second];
    }
    for (int i = 1; i < n; i++) cout << sol[i] << " ";
  } else
    cout << -1;
  return 0;
}