#include <bits/stdc++.h>
using namespace std;
short int n, m, nod[5000 + 10][5000 + 10], nivel[5000 + 10], tata[5000 + 10];
int sol[5000 + 10];
bool viz[5000 + 10];
struct query {
  int x, y, c;
} q[5000 + 10];
map<pair<short int, short int>, short int> umap;
vector<pair<short int, short int> > a;
void dfs(int x) {
  viz[x] = 1;
  for (int i = 1; i <= n; i++)
    if (nod[x][i] && !viz[i]) {
      nivel[i] = nivel[x] + 1;
      tata[i] = x;
      dfs(i);
    }
}
inline bool mycmp(query a, query b) {
  if (a.c < b.c) return 1;
  if (a.c == b.c && a.x < b.x) return 1;
  if (a.c == b.c && a.x == b.x && a.y < b.y) return 1;
  return 0;
}
void LCA(int x1, int x2, int x3) {
  if (nivel[x1] < nivel[x2]) swap(x1, x2);
  while (nivel[x1] > nivel[x2]) {
    int t = tata[x1];
    if (nod[x1][t]) a.push_back(make_pair(x1, t));
    x1 = t;
  }
  while (x1 != x2) {
    int t1 = tata[x1], t2 = tata[x2];
    if (nod[x1][t1]) a.push_back(make_pair(x1, t1));
    if (nod[x2][t2]) a.push_back(make_pair(x2, t2));
    x1 = t1;
    x2 = t2;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int nod1, nod2;
    cin >> nod1 >> nod2;
    umap[make_pair(nod1, nod2)] = umap[make_pair(nod2, nod1)] = i;
    sol[i] = 1000000;
    nod[nod1][nod2] = nod[nod2][nod1] = 1;
  }
  nivel[1] = 1;
  dfs(1);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> q[i].x >> q[i].y >> q[i].c;
    if (q[i].x > q[i].y) swap(q[i].x, q[i].y);
  }
  sort(q + 1, q + m + 1, mycmp);
  int i = m;
  while (i) {
    a.clear();
    int val = q[i].c;
    while (i && q[i].c == val) {
      if (q[i].x == q[i + 1].x && q[i].y == q[i + 1].y &&
          q[i].c == q[i + 1].c) {
        i--;
        continue;
      }
      int cnt = a.size();
      LCA(q[i].x, q[i].y, q[i].c);
      if (a.size() == cnt) {
        cout << -1 << '\n';
        return 0;
      }
      i--;
    }
    for (int i = 0; i < a.size(); i++) {
      nod[a[i].first][a[i].second] = nod[a[i].first][a[i].second] = 0;
      sol[umap[a[i]]] = val;
    }
  }
  for (int i = 1; i < n; i++) cout << sol[i] << ' ';
  cout << '\n';
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
short int n, m, nod[5000 + 10][5000 + 10], nivel[5000 + 10], tata[5000 + 10];
int sol[5000 + 10];
bool viz[5000 + 10];
struct query {
  int x, y, c;
} q[5000 + 10];
map<pair<short int, short int>, short int> umap;
vector<pair<short int, short int> > a;
void dfs(int x) {
  viz[x] = 1;
  for (int i = 1; i <= n; i++)
    if (nod[x][i] && !viz[i]) {
      nivel[i] = nivel[x] + 1;
      tata[i] = x;
      dfs(i);
    }
}
inline bool mycmp(query a, query b) {
  if (a.c < b.c) return 1;
  if (a.c == b.c && a.x < b.x) return 1;
  if (a.c == b.c && a.x == b.x && a.y < b.y) return 1;
  return 0;
}
void LCA(int x1, int x2, int x3) {
  if (nivel[x1] < nivel[x2]) swap(x1, x2);
  while (nivel[x1] > nivel[x2]) {
    int t = tata[x1];
    if (nod[x1][t]) a.push_back(make_pair(x1, t));
    x1 = t;
  }
  while (x1 != x2) {
    int t1 = tata[x1], t2 = tata[x2];
    if (nod[x1][t1]) a.push_back(make_pair(x1, t1));
    if (nod[x2][t2]) a.push_back(make_pair(x2, t2));
    x1 = t1;
    x2 = t2;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int nod1, nod2;
    cin >> nod1 >> nod2;
    umap[make_pair(nod1, nod2)] = umap[make_pair(nod2, nod1)] = i;
    sol[i] = 1000000;
    nod[nod1][nod2] = nod[nod2][nod1] = 1;
  }
  nivel[1] = 1;
  dfs(1);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> q[i].x >> q[i].y >> q[i].c;
    if (q[i].x > q[i].y) swap(q[i].x, q[i].y);
  }
  sort(q + 1, q + m + 1, mycmp);
  int i = m;
  while (i) {
    a.clear();
    int val = q[i].c;
    while (i && q[i].c == val) {
      if (q[i].x == q[i + 1].x && q[i].y == q[i + 1].y &&
          q[i].c == q[i + 1].c) {
        i--;
        continue;
      }
      int cnt = a.size();
      LCA(q[i].x, q[i].y, q[i].c);
      if (a.size() == cnt) {
        cout << -1 << '\n';
        return 0;
      }
      i--;
    }
    for (int i = 0; i < a.size(); i++) {
      nod[a[i].first][a[i].second] = nod[a[i].first][a[i].second] = 0;
      sol[umap[a[i]]] = val;
    }
  }
  for (int i = 1; i < n; i++) cout << sol[i] << ' ';
  cout << '\n';
  return 0;
}

```