#include <bits/stdc++.h>
using namespace std;
const int N = 5005;
vector<int> g[N];
pair<int, int> edge[N];
int val[N], dep[N], par[N];
void dfs(int v, int p, int __dep = 0) {
  dep[v] = __dep;
  par[v] = p;
  for (auto x : g[v]) {
    if (x == p) continue;
    if (edge[x].first == v)
      dfs(edge[x].second, x, __dep + 1);
    else
      dfs(edge[x].first, x, __dep + 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(32);
  int n;
  cin >> n;
  int u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    edge[i] = {u, v};
    g[u].push_back(i);
    g[v].push_back(i);
    val[i] = 1;
  }
  dfs(1, 0);
  int m, a, b, g;
  cin >> m;
  vector<pair<pair<int, int>, int> > vec;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> g;
    vec.push_back({{a, b}, g});
    if (dep[a] < dep[b]) swap(a, b);
    while (dep[a] > dep[b]) {
      int x = par[a];
      val[x] = max(val[x], g);
      a = edge[x].first + edge[x].second - a;
    }
    while (a != b) {
      int x = par[a];
      val[x] = max(val[x], g);
      a = edge[x].first + edge[x].second - a;
      x = par[b];
      val[x] = max(val[x], g);
      b = edge[x].first + edge[x].second - b;
    }
  }
  for (int i = 0; i < m; i++) {
    a = vec[i].first.first;
    b = vec[i].first.second;
    g = vec[i].second;
    int mini = 1000000;
    if (dep[a] < dep[b]) swap(a, b);
    while (dep[a] > dep[b]) {
      int x = par[a];
      mini = min(val[x], mini);
      a = edge[x].first + edge[x].second - a;
    }
    while (a != b) {
      int x = par[a];
      mini = min(val[x], mini);
      a = edge[x].first + edge[x].second - a;
      x = par[b];
      mini = min(val[x], mini);
      b = edge[x].first + edge[x].second - b;
    }
    if (mini != g) {
      cout << -1 << endl;
      exit(0);
    }
  }
  for (int i = 1; i < n; i++) {
    cout << val[i] << " ";
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 5005;
vector<int> g[N];
pair<int, int> edge[N];
int val[N], dep[N], par[N];
void dfs(int v, int p, int __dep = 0) {
  dep[v] = __dep;
  par[v] = p;
  for (auto x : g[v]) {
    if (x == p) continue;
    if (edge[x].first == v)
      dfs(edge[x].second, x, __dep + 1);
    else
      dfs(edge[x].first, x, __dep + 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(32);
  int n;
  cin >> n;
  int u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    edge[i] = {u, v};
    g[u].push_back(i);
    g[v].push_back(i);
    val[i] = 1;
  }
  dfs(1, 0);
  int m, a, b, g;
  cin >> m;
  vector<pair<pair<int, int>, int> > vec;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> g;
    vec.push_back({{a, b}, g});
    if (dep[a] < dep[b]) swap(a, b);
    while (dep[a] > dep[b]) {
      int x = par[a];
      val[x] = max(val[x], g);
      a = edge[x].first + edge[x].second - a;
    }
    while (a != b) {
      int x = par[a];
      val[x] = max(val[x], g);
      a = edge[x].first + edge[x].second - a;
      x = par[b];
      val[x] = max(val[x], g);
      b = edge[x].first + edge[x].second - b;
    }
  }
  for (int i = 0; i < m; i++) {
    a = vec[i].first.first;
    b = vec[i].first.second;
    g = vec[i].second;
    int mini = 1000000;
    if (dep[a] < dep[b]) swap(a, b);
    while (dep[a] > dep[b]) {
      int x = par[a];
      mini = min(val[x], mini);
      a = edge[x].first + edge[x].second - a;
    }
    while (a != b) {
      int x = par[a];
      mini = min(val[x], mini);
      a = edge[x].first + edge[x].second - a;
      x = par[b];
      mini = min(val[x], mini);
      b = edge[x].first + edge[x].second - b;
    }
    if (mini != g) {
      cout << -1 << endl;
      exit(0);
    }
  }
  for (int i = 1; i < n; i++) {
    cout << val[i] << " ";
  }
  return 0;
}

```