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;
pair<int, int> edge[5001];
int u[5001], v[5001], g[5001];
vector<int> con[5001];
vector<int> son[5001];
bool vis[5001];
int father[5001];
int level[5001];
bool ok = true;
int ans[5001];
int ans2[5001];
void wk(int x, int y, int val) {
  if (x == y) {
    return;
  } else if (level[x] < level[y]) {
    ans[y] = max(ans[y], val);
    wk(x, father[y], val);
  } else {
    ans[x] = max(ans[x], val);
    wk(father[x], y, val);
  }
}
void ck(int x, int y, int val, bool fd) {
  if (x == y) {
    if (!fd) ok = false;
    return;
  } else if (level[x] < level[y]) {
    if (ans[y] < val) ok = false;
    if (ans[y] == val) fd = true;
    ck(x, father[y], val, fd);
  } else {
    if (ans[x] < val) ok = false;
    if (ans[x] == val) fd = true;
    ck(father[x], y, val, fd);
  }
}
void fd(int x) {
  int i;
  for (i = 1; i < n; i++) {
    if (edge[i] == pair<int, int>(x, father[x]) ||
        edge[i] == pair<int, int>(father[x], x)) {
      ans2[i] = ans[x];
      break;
    }
  }
}
void root(int x, int lev) {
  int i;
  vis[x] = true, level[x] = lev;
  for (i = 0; i < con[x].size(); i++) {
    int f = con[x][i];
    if (!vis[f])
      vis[f] = true, son[x].push_back(f), father[f] = x, root(f, lev + 1);
  }
}
int main() {
  int i;
  cin >> n;
  for (i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    edge[i] = pair<int, int>(x, y);
    con[x].push_back(y), con[y].push_back(x);
  }
  cin >> m;
  for (i = 1; i <= m; i++) cin >> u[i] >> v[i] >> g[i];
  root(1, 1);
  for (i = 1; i <= m; i++) wk(u[i], v[i], g[i]);
  for (i = 1; i <= m; i++) ck(u[i], v[i], g[i], false);
  for (i = 2; i <= n; i++) fd(i);
  for (i = 1; i <= n; i++)
    if (ans2[i] == 0) ans2[i] = 1000000;
  if (ok) {
    for (i = 1; i < n - 1; i++) cout << ans2[i] << " ";
    cout << ans2[n - 1] << endl;
  } else
    cout << -1 << endl;
  return 0;
}