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 INF = 0x3f3f3f3f;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
const int N = 5005;
vector<pair<int, int>> G[N];
int col[N], id[N], ans[N];
int fa[N], dep[N];
void dfs(int u, int f) {
  dep[u] = dep[f] + 1;
  fa[u] = f;
  for (auto it : G[u]) {
    if (it.first == f) continue;
    id[it.first] = it.second;
    dfs(it.first, u);
  }
}
void update(int x, int y, int z) {
  while (x != y) {
    if (dep[x] < dep[y]) swap(x, y);
    col[x] = max(col[x], z);
    x = fa[x];
  }
}
int query(int x, int y) {
  int mn = 1e7;
  while (x != y) {
    if (dep[x] < dep[y]) swap(x, y);
    mn = min(mn, col[x]);
    x = fa[x];
  }
  return mn;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 1; i <= int(n - 1); ++i) {
    int x, y;
    cin >> x >> y;
    G[x].emplace_back(y, i);
    G[y].emplace_back(x, i);
  }
  dfs(1, 0);
  int m;
  cin >> m;
  vector<tuple<int, int, int>> q(m);
  for (int i = 0; i < int(m); ++i) {
    int x, y, z;
    cin >> x >> y >> z;
    update(x, y, z);
    q[i] = make_tuple(x, y, z);
  }
  for (auto it : q) {
    int x, y, z;
    tie(x, y, z) = it;
    if (query(x, y) != z) {
      cout << -1;
      return 0;
    }
  }
  for (int i = int(2); i <= int(n); ++i) {
    if (col[i])
      ans[id[i]] = col[i];
    else
      ans[id[i]] = 1e6;
  }
  for (int i = 1; i <= int(n - 1); ++i) cout << ans[i] << ' ';
  return 0;
}